CBSE Class 10 Science Light reflection and refraction Notes

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CHAPTER 10: LIGHT – REFLECTION AND REFRACTION

LIGHT

An object reflects light that falls on it. This reflected light, when received by our eyes, enables us to see things.

Reflection of light

Reflection of light is the phenomenon of bouncing back of light in the same medium on striking the surface of any object.

There are two types of reflection:

1. Regular reflection or specular reflection
2. Irregular reflection or diffuse reflection

Regular reflection: When the reflecting surface is smooth and well polished, the parallel rays falling on it are reflected parallel to one another; the reflected light goes in one particular direction. This is regular reflection or specular reflection.

Irregular reflection: When the reflecting surface is rough, the parallel rays falling on it are reflected in different directions. Such a reflection is known as diffuse reflection or irregular reflection.

LAWS OF REFLECTION OF LIGHT

According to the laws of reflection of light,

1. The angle of incidence is equal to the angle of reflection, and
2. The incident ray, the normal to the mirror at the point of incidence and the reflected ray all lie in the same plane.

These laws of reflection are applicable to all types of reflecting surfaces, including spherical surfaces.

OBJECTS

Anything which gives out light rays, either its own or reflected by it, is called an object.

Luminous objects: The objects like the Sun, other stars, an electric bulb, a tube-light, etc. which emit their own light are called luminous objects.

Non-luminous objects: The objects which do not emit light themselves but only reflect or scatter the light which falls on them are called non-luminous objects. A flower, chair, table, book, trees, etc. are all non-luminous objects.

IMAGES

An image is an optical appearance produced when light rays coming from an object are reflected from a mirror (or refracted through a lens).

Real image: The image which can be obtained on a screen is called a real image. In a cinema hall, we see the images of actors and actresses on the screen. So, the images formed on a cinema screen are examples of real images.

Virtual image: The image which cannot be obtained on a screen is called a virtual image. A virtual image can be seen only by looking into a mirror. The image of our face in a plane mirror is an example of a virtual image.

LATERAL INVERSION

When an object is placed in front of a plane mirror, the right side of the object appears to become the left side of the image, and the left side of the object appears to become the right side of the image. This change of sides of an object and its mirror image is called lateral inversion. The phenomenon of lateral inversion is due to the reflection of light.

CHARACTERISTICS OF IMAGES FORMED BY PLANE MIRRORS

1. The image of a real object is always virtual. Such an image cannot be taken on a screen.
2. The image formed in a plane mirror is always erect.
3. The size of the image in a plane mirror is always the same as the size of the object.
4. The image formed in a plane mirror is as far behind the mirror as the object is in front of the mirror.
5. The image formed in a plane mirror is laterally inverted, i.e. the left side of the object becomes the right side of the image and vice-versa.

SPHERICAL MIRROR

A spherical mirror is a mirror whose reflecting surface is a part of a hollow sphere of glass. Spherical mirrors are of two types: concave mirror and convex mirror.

Concave mirror: A concave mirror is a spherical mirror in which the reflection of light takes place at the concave surface (or bent-in surface).

Convex mirror: A convex mirror is a spherical mirror in which the reflection of light takes place at the convex surface (or bulging-out surface).

TERMS RELATED TO SPHERICAL MIRRORS

Centre of curvature (C): The centre of curvature of a spherical mirror is the centre of the hollow sphere of glass of which the spherical mirror is a part. It is represented by the letter C.

Pole (P): The pole of a spherical mirror is the centre of the mirror. It is represented by the letter P.

Radius of curvature (R): The radius of curvature of a spherical mirror is the radius of the hollow sphere of glass of which the spherical mirror is a part. It is represented by the letter R.

Principal axis: The principal axis of a spherical mirror is the straight line passing through the centre of curvature C and the pole P of the spherical mirror, produced on both sides.

Aperture: The aperture of a spherical mirror is the diameter of the reflecting surface of the mirror.

PRINCIPAL FOCUS OF A SPHERICAL MIRROR

The principal focus of a concave mirror is a point on its principal axis to which all the light rays which are parallel and close to the axis converge after reflection from the concave mirror. A concave mirror has a real focus. The focus of a concave mirror is in front of the mirror. Since a concave mirror converges a parallel beam of light rays, it is also called a converging mirror.

The principal focus of a convex mirror is a point on its principal axis from which a beam of light rays, initially parallel to the axis, appears to diverge after being reflected from the convex mirror. A convex mirror has a virtual focus. The focus of a convex mirror is situated behind the mirror. Since a convex mirror diverges a parallel beam of light rays, it is also called a diverging mirror.

Focal length: The focal length of a spherical mirror is the distance between its pole and principal focus. It is denoted by the letter f.

Relation between radius of curvature and focal length of a spherical mirror

The focal length of a spherical mirror is equal to half of its radius of curvature.

\[f=\dfrac{R}{2}\]

In other words, for spherical mirrors of small apertures, the radius of curvature is equal to twice the focal length: \(R=2f\).

RULES FOR OBTAINING IMAGES FORMED BY SPHERICAL MIRRORS

The intersection of at least two reflected rays gives the position of the image of the point object. Any two of the following rays can be considered for locating the image.

1. A ray parallel to the principal axis, after reflection, will pass through the principal focus in the case of a concave mirror, or appear to diverge from the principal focus in the case of a convex mirror.

2. A ray passing through the principal focus of a concave mirror, or a ray which is directed towards the principal focus of a convex mirror, after reflection, will emerge parallel to the principal axis.

3. A ray passing through the centre of curvature of a concave mirror, or directed in the direction of the centre of curvature of a convex mirror, after reflection, is reflected back along the same path. The light rays come back along the same path because the incident rays fall on the mirror along the normal to the reflecting surface.

4. A ray incident obliquely to the principal axis, towards the point P (pole of the mirror) on a concave or a convex mirror, is reflected obliquely. The incident and reflected rays follow the laws of reflection at the point of incidence (point P), making equal angles with the principal axis.

FORMATION OF DIFFERENT TYPES OF IMAGES BY A CONCAVE MIRROR

The type of image formed by a concave mirror depends on the position of the object in front of the mirror. There are six positions of the object:

Case 1: Object is in between P and F

1. behind the mirror
2. virtual and erect, and
3. larger than the object (or magnified).

Case 2: Object is at the focus (F)

1. at infinity
2. real and inverted, and
3. highly magnified (or highly enlarged).

Case 3: Object is in between the focus (F) and the centre of curvature (C)

1. beyond the centre of curvature
2. real and inverted, and
3. larger than the object (or magnified).

Case 4: Object is at the centre of curvature (C)

1. at the centre of curvature
2. real and inverted, and
3. of the same size as the object.

Case 5: Object is beyond the centre of curvature (C)

1. between the focus and the centre of curvature
2. real and inverted, and
3. smaller than the object (or diminished).

Case 6: Object is at infinity

1. at the focus F
2. real and inverted, and
3. much smaller than the object (or highly diminished).

USES OF CONCAVE MIRRORS

1. Concave mirrors are commonly used in torches, search-lights and vehicle headlights to get powerful parallel beams of light.
2. Concave mirrors are used as shaving mirrors to see a larger image of the face.
3. Dentists use concave mirrors to see large images of the teeth of patients.
4. Concave mirrors are used as doctors’ head mirrors to focus light coming from a lamp on to the body parts of a patient to be examined by the doctor.
5. Concave dishes are used in TV dish antennas to receive TV signals from the distant communications satellite.
6. Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

FORMATION OF DIFFERENT TYPES OF IMAGES BY A CONVEX MIRROR

The type of image formed by a convex mirror depends on the position of the object in front of the mirror. There are two positions of the object:

Case 1: Object is placed between P and infinity

1. between the pole and the focus (behind the mirror)
2. virtual and erect, and
3. smaller than the object (or diminished).

Case 2: Object is at infinity

1. behind the mirror at the focus
2. virtual and erect, and
3. much smaller than the object (or highly diminished).

USES OF CONVEX MIRRORS

Convex mirrors are commonly used as rear-view (wing) mirrors in vehicles. These mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving. Convex mirrors are preferred because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view a much larger area than would be possible with a plane mirror.

INTEXT QUESTIONS (PAGE 168)

Define the principal focus of a concave mirror.

Answer: Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.

The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer: Here \(R=20\,\mathrm{cm}\).
\(f=\dfrac{R}{2}=\dfrac{20}{2}=10\,\mathrm{cm}\)

Name a mirror that can give an erect and enlarged image of an object.

Answer: A concave mirror. When an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect and enlarged.

Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer: Convex mirrors give a virtual, erect and diminished image of the objects placed in front of them. They are preferred as rear-view mirrors in vehicles because they give a wider field of view, which allows the driver to see most of the traffic behind him.

MIRROR FORMULA

In a spherical mirror, the distance of the object from its pole is called the object distance (u). The distance of the image from the pole of the mirror is called the image distance (v). The distance of the principal focus from the pole is called the focal length (f). There is a relationship between these three quantities, given by the mirror formula:

\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]

MAGNIFICATION

Magnification produced by a spherical mirror gives the relative extent to which the image of an object is magnified with respect to the object size. It is expressed as the ratio of the height of the image to the height of the object. It is usually represented by the letter m. If \(h_1\) is the height of the object and \(h_2\) is the height of the image, then the magnification m produced by a spherical mirror is given by

\[m=\dfrac{\text{height of the image}}{\text{height of the object}}=\dfrac{h_2}{h_1}\]

The magnification m is also related to the object distance (u) and the image distance (v). It can be expressed as

\[m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}\]

Points to be remembered

• The height of the object is taken to be positive as the object is usually placed above the principal axis.
• The height of the image should be taken as positive for virtual images. However, it is to be taken as negative for real images.
• When the image is real, it is inverted, so \(h_2\) is negative, which results in m being negative. A negative sign in the value of the magnification indicates that the image is real.
• When the image is virtual, it is erect, so \(h_2\) is positive, which results in m being positive. A positive sign in the value of the magnification indicates that the image is virtual.

SIGN CONVENTION FOR SPHERICAL MIRRORS

The following sign convention is used for measuring various distances in the ray diagrams of spherical mirrors:

1. The object is always placed to the left of the mirror.
2. All distances are measured from the pole of the mirror.
3. Distances measured in the direction of the incident ray are positive, and distances measured in the direction opposite to that of the incident rays are negative.
4. Distances measured above the principal axis are positive, and those measured below the principal axis are negative.

As a result, the focal length of a concave mirror is negative and that of a convex mirror is positive; the object distance u is always negative.

INTEXT QUESTIONS (PAGE 171)

Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer: Here \(R=32\,\mathrm{cm}\).
\(f=\dfrac{R}{2}=\dfrac{32}{2}=16\,\mathrm{cm}\)
Hence, the focal length of the given convex mirror is 16 cm.

A concave mirror produces a three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer: Here, magnification \(m=-3\), object distance \(u=-10\,\mathrm{cm}\) and image distance v = ?
Putting these values in the magnification formula for a mirror,
\(m=-\dfrac{v}{u}\;\Rightarrow\;-3=-\dfrac{v}{-10}\;\Rightarrow\;v=-30\,\mathrm{cm}\)
The image is formed 30 cm in front of the mirror.

NUMERICALS BASED ON CONVEX AND CONCAVE MIRRORS

Find the focal length of a convex mirror of radius of curvature 1 m.

Answer: \(f=\dfrac{R}{2}=\dfrac{100}{2}=50\,\mathrm{cm}\) (= 0.5 m, positive for a convex mirror).

The focal length of a convex mirror is 50 cm. What is its radius of curvature?

Answer: \(R=2f=2\times50=100\,\mathrm{cm}\) (= 1 m).

The radius of curvature of a concave mirror is 25 cm. What is its focal length?

Answer: \(f=\dfrac{R}{2}=\dfrac{25}{2}=12.5\,\mathrm{cm}\). By the sign convention, \(f=-12.5\,\mathrm{cm}\) for the concave mirror.

A concave mirror produces a 10 cm long image of an object of height 2 cm. What is the magnification produced?

Answer: \(m=\dfrac{h_2}{h_1}=\dfrac{-10}{2}=-5\)
The magnitude of the magnification is 5; the negative sign shows that the image is real and inverted (a real image is formed by a concave mirror).

An object 1 cm high is held near a concave mirror of magnification 10. How tall will the image be?

Answer: \(h_2=m\,h_1=10\times1=10\,\mathrm{cm}\)
The image is 10 cm tall.

An object 4 cm in size is placed at a distance of 25 cm from a concave mirror of focal length 15 cm. Find the position, nature and height of the image.

Answer: \(u=-25\,\mathrm{cm},\; f=-15\,\mathrm{cm},\; h_1=4\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{1}{15}+\dfrac{1}{25}=\dfrac{-5+3}{75}=-\dfrac{2}{75}\;\Rightarrow\;v=-37.5\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=-\dfrac{-37.5}{-25}=-1.5\;\Rightarrow\;h_2=m\,h_1=-6\,\mathrm{cm}\)
The image is formed 37.5 cm in front of the mirror; it is real, inverted and magnified, 6 cm in height.

A converging mirror forms a real image of height 4 cm, of an object of height 1 cm placed 20 cm away from the mirror. Calculate the image distance. What is the focal length of the mirror?

Answer: \(h_1=1\,\mathrm{cm},\; h_2=-4\,\mathrm{cm},\; u=-20\,\mathrm{cm}\)
\(m=\dfrac{h_2}{h_1}=-4=-\dfrac{v}{u}\;\Rightarrow\;v=4u=-80\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{1}{80}-\dfrac{1}{20}=-\dfrac{5}{80}\;\Rightarrow\;f=-16\,\mathrm{cm}\)
The image is 80 cm in front of the mirror and the focal length is 16 cm (concave mirror).

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Answer: \(u=-12\,\mathrm{cm},\; f=+15\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{15}+\dfrac{1}{12}=\dfrac{9}{60}\;\Rightarrow\;v=+6.67\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=\dfrac{6.67}{12}=+0.56\)
The image is 6.67 cm behind the mirror, virtual, erect and diminished (height 2.5 cm). As the needle is moved farther from the mirror, the image moves towards the focus (15 cm behind the mirror) and becomes smaller and smaller, but it always remains virtual and erect.

An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.

Answer: \(u=-25\,\mathrm{cm},\; f=+20\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{25}=\dfrac{9}{100}\;\Rightarrow\;v=+11.1\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=\dfrac{11.1}{25}=+0.44\;\Rightarrow\;h_2=0.44\times2.5=1.1\,\mathrm{cm}\)
The image is virtual, erect and diminished, formed 11.1 cm behind the mirror; its size is about 1.1 cm.

The image formed by a convex mirror of focal length 20 cm is a quarter of the object. What is the distance of the object from the mirror?

Answer: \(m=+\dfrac14=-\dfrac{v}{u}\;\Rightarrow\;v=-\dfrac{u}{4}\)
\(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{4}{u}+\dfrac{1}{u}=-\dfrac{3}{u}\;\Rightarrow\;u=-3f=-60\,\mathrm{cm}\)
The object is 60 cm in front of the mirror.

Find the size, nature and position of the image formed by a concave mirror, when an object of size 1 cm is placed at a distance of 15 cm. The focal length of the mirror is 10 cm.

Answer: \(u=-15\,\mathrm{cm},\; f=-10\,\mathrm{cm}\)
\(\dfrac{1}{v}=-\dfrac{1}{10}+\dfrac{1}{15}=-\dfrac{1}{30}\;\Rightarrow\;v=-30\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=-\dfrac{-30}{-15}=-2\;\Rightarrow\;h_2=-2\,\mathrm{cm}\)
The image is real and inverted, magnified to 2 cm, and formed 30 cm in front of the mirror.

An object 2 cm high is placed at a distance of 16 cm from a concave mirror, which produces a 3 cm high inverted image. What is the focal length of the mirror? Also find the position of the image.

Answer: \(m=\dfrac{h_2}{h_1}=\dfrac{-3}{2}=-1.5=-\dfrac{v}{u}\;\Rightarrow\;v=1.5u=-24\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{1}{24}-\dfrac{1}{16}=-\dfrac{5}{48}\;\Rightarrow\;f=-9.6\,\mathrm{cm}\)
The focal length is 9.6 cm (concave) and the image is formed 24 cm in front of the mirror.

An erect image 3 times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. What is the position of the object?

Answer: \(f=-\dfrac{36}{2}=-18\,\mathrm{cm},\quad m=+3=-\dfrac{v}{u}\;\Rightarrow\;v=-3u\)
\(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{1}{3u}+\dfrac{1}{u}=\dfrac{2}{3u}\;\Rightarrow\;u=\dfrac{2f}{3}=-12\,\mathrm{cm}\)
The object is 12 cm in front of the mirror (between the pole and the focus).

A 2.5 cm candle is placed 12 cm away from a convex mirror of focal length 30 cm. Give the location of the image and the magnification.

Answer: \(\dfrac{1}{v}=\dfrac{1}{30}+\dfrac{1}{12}=\dfrac{7}{60}\;\Rightarrow\;v=+8.57\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=\dfrac{8.57}{12}=+0.71\)
The image is virtual and erect, 8.57 cm behind the mirror, with magnification +0.71 (image height about 1.8 cm).

An object is placed in front of a concave mirror of focal length 20 cm. The image formed is 3 times the size of the object. Calculate the two possible distances of the object from the mirror.

Answer: \(f=-20\,\mathrm{cm}\)
Real image (\(m=-3\)): \(v=3u,\;\dfrac{1}{f}=\dfrac{1}{3u}+\dfrac{1}{u}=\dfrac{4}{3u}\;\Rightarrow\;u=\dfrac{4f}{3}=-26.7\,\mathrm{cm}\)
Virtual image (\(m=+3\)): \(v=-3u,\;\dfrac{1}{f}=-\dfrac{1}{3u}+\dfrac{1}{u}=\dfrac{2}{3u}\;\Rightarrow\;u=\dfrac{2f}{3}=-13.3\,\mathrm{cm}\)
The object may be 26.7 cm or 13.3 cm from the mirror.

The image formed by a convex mirror is virtual, erect and smaller in size. Illustrate with a figure.

Answer: See the figure below: the object is in front of the convex mirror and the image is formed behind the mirror, between the pole and the focus.

A concave mirror produces a real image 10 mm tall, of an object 2.5 mm tall placed at 5 cm from the mirror. Calculate the focal length of the mirror and the position of the image.

Answer: \(m=\dfrac{-10}{2.5}=-4=-\dfrac{v}{u}\;\Rightarrow\;v=4u=-20\,\mathrm{cm}\)
\(\dfrac{1}{f}=-\dfrac{1}{20}-\dfrac{1}{5}=-\dfrac{1}{4}\;\Rightarrow\;f=-4\,\mathrm{cm}\)
The focal length is 4 cm and the image is formed 20 cm in front of the mirror.

An object is placed at a large distance in front of a convex mirror of radius of curvature 40 cm. How far is the image behind the mirror?

Answer: \(f=\dfrac{R}{2}=20\,\mathrm{cm}\)
For an object at a very large distance the image is formed at the principal focus, so the image is 20 cm behind the mirror.

An object is placed 15 cm from a convex mirror of radius of curvature 90 cm. Calculate the position of the image and its magnification.

Answer: \(f=\dfrac{90}{2}=+45\,\mathrm{cm},\quad u=-15\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{45}+\dfrac{1}{15}=\dfrac{4}{45}\;\Rightarrow\;v=+11.25\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=\dfrac{11.25}{15}=+0.75\)
The image is 11.25 cm behind the mirror, virtual and erect, with magnification +0.75.

The image formed by a convex mirror of focal length 30 cm is a quarter of the object. What is the distance of the object from the mirror?

Answer: \(u=-3f=-3\times30=-90\,\mathrm{cm}\)
(Same working as in the earlier problem with \(m=+\tfrac14\).) The object is 90 cm in front of the mirror.

When an object is placed at a distance of 60 cm from a convex mirror, the magnification produced is 1/2. Where should the object be placed to get a magnification of 1/3?

Answer: \(m=\dfrac{f}{f-u}\)
For \(m=\tfrac12\) and \(u=-60\,\mathrm{cm}\): \(\dfrac{f}{f+60}=\dfrac12\;\Rightarrow\;f=60\,\mathrm{cm}\)
For \(m=\tfrac13\): \(\dfrac{60}{60-u}=\dfrac13\;\Rightarrow\;60-u=180\;\Rightarrow\;u=-120\,\mathrm{cm}\)
The object should be placed 120 cm in front of the mirror.

An object is placed 18 cm in front of a mirror. If the image is formed 4 cm to the right of the mirror, calculate its focal length. Is the mirror convex or concave? What is the nature of the image? What is the radius of curvature of the mirror?

Answer: \(u=-18\,\mathrm{cm},\; v=+4\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=\dfrac14-\dfrac{1}{18}=\dfrac{7}{36}\;\Rightarrow\;f=+5.14\,\mathrm{cm}\)
The focal length is positive, so the mirror is convex. The image is behind the mirror, so it is virtual and erect (\(m=-\dfrac{4}{-18}=+0.22\), diminished). \(R=2f=10.3\,\mathrm{cm}\).

A convex mirror used for rear view on an automobile has a radius of curvature of 3 m. If a bus is located at 5 m from this mirror, find the position, nature and magnification of the image.

Answer: \(f=\dfrac{3}{2}=1.5\,\mathrm{m},\quad u=-5\,\mathrm{m}\)
\(\dfrac{1}{v}=\dfrac{1}{1.5}+\dfrac{1}{5}=\dfrac{13}{15}\;\Rightarrow\;v=+1.15\,\mathrm{m}\)
\(m=-\dfrac{v}{u}=\dfrac{1.15}{5}=+0.23\)
The image is 1.15 m behind the mirror; it is virtual, erect and diminished.

An object 3 cm high is held at a distance of 50 cm from a diverging mirror of focal length 25 cm. Find the nature, position and size of the image formed.

Answer: \(\dfrac{1}{v}=\dfrac{1}{25}+\dfrac{1}{50}=\dfrac{3}{50}\;\Rightarrow\;v=+16.7\,\mathrm{cm}\)
\(m=-\dfrac{v}{u}=\dfrac{16.7}{50}=+0.33\;\Rightarrow\;h_2=1\,\mathrm{cm}\)
The image is virtual, erect and diminished, 16.7 cm behind the mirror, and 1 cm in size.

A converging mirror of focal length 20 cm forms an image which is two times the size of the object. Calculate the two possible distances of the object from the mirror.

Answer: \(f=-20\,\mathrm{cm}\)
Real image (\(m=-2\)): \(v=2u,\;\dfrac{1}{f}=\dfrac{3}{2u}\;\Rightarrow\;u=\dfrac{3f}{2}=-30\,\mathrm{cm}\)
Virtual image (\(m=+2\)): \(v=-2u,\;\dfrac{1}{f}=\dfrac{1}{2u}\;\Rightarrow\;u=\dfrac{f}{2}=-10\,\mathrm{cm}\)
The object may be 30 cm or 10 cm from the mirror.

The linear magnification of a convex mirror of focal length 15 cm is 1/3. What is the distance of the object from the focus of the mirror?

Answer: \(m=\dfrac{f}{f-u}=\dfrac13\;\Rightarrow\;f-u=45\;\Rightarrow\;u=-30\,\mathrm{cm}\)
The object is 30 cm in front of the mirror. The focus of a convex mirror lies 15 cm behind the mirror, so the object is \(30+15=45\,\mathrm{cm}\) from the focus.

The focal length of a convex mirror is 12.5 cm. How far is its centre of curvature (i) from the pole, (ii) from the focus?

Answer: \(R=2f=25\,\mathrm{cm}\)
(i) The centre of curvature is 25 cm from the pole (behind the mirror).
(ii) The focus is 12.5 cm from the pole, so the centre of curvature is \(25-12.5=12.5\,\mathrm{cm}\) from the focus.

Find the focal length of a concave mirror that produces a four times larger real image of an object held at 5 cm from the mirror.

Answer: \(m=-4\;\Rightarrow\;v=4u=-20\,\mathrm{cm}\)
\(\dfrac{1}{f}=-\dfrac{1}{20}-\dfrac{1}{5}=-\dfrac{1}{4}\;\Rightarrow\;f=-4\,\mathrm{cm}\)
The focal length of the concave mirror is 4 cm.

An object is held at 30 cm in front of a convex mirror of focal length 15 cm. At what distance from the convex mirror should a plane mirror be held so that the two images coincide with each other?

Answer: \(\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{30}=\dfrac{1}{10}\;\Rightarrow\;v=+10\,\mathrm{cm}\)
The convex mirror forms the image 10 cm behind it. Let the plane mirror be held d cm in front of the convex mirror. The object is then (30 − d) cm in front of the plane mirror, so its image is (30 − d) cm behind the plane mirror, i.e. at a distance \((30-d)-d=30-2d\) behind the convex mirror.
For the images to coincide: \(30-2d=10\;\Rightarrow\;d=10\,\mathrm{cm}\).
The plane mirror should be held 10 cm in front of the convex mirror.

Draw any three ray diagrams to show how the size and nature of the image of an object change when it moves from the centre of curvature of a concave mirror towards the pole of the mirror.

Answer: Object at C: image at C, real, inverted, same size.

REFRACTION OF LIGHT

The change in direction of light when it passes from one medium to another obliquely is called refraction of light. In other words, the bending of light when it goes from one medium to another obliquely is called refraction of light. Refraction takes place when light enters from air into water.

The speed of light is different in different substances. The refraction of light is due to the change in the speed of light on going from one medium to another. Thus, when light goes from one medium to another, its speed changes, and this change in speed of light causes the refraction of light.

MEDIUM

A transparent substance in which light travels is known as a medium. Media can be divided into two types:

1. Optically rarer medium: A medium in which the speed of light is more is known as an optically rarer medium (or less dense medium).
2. Optically denser medium: A medium in which the speed of light is less is known as an optically denser medium (or more dense medium).

Glass is an optically denser medium than air and water.

RULES OF REFRACTION

Rule 1: When a light ray travels from a rarer medium to a denser medium, the light ray bends towards the normal.

Rule 2: When a light ray travels from a denser medium to a rarer medium, the light ray bends away from the normal.

LAWS OF REFRACTION

According to the laws of refraction of light,

1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell’s law of refraction.

If i is the angle of incidence and r is the angle of refraction, then

\[\dfrac{\sin i}{\sin r}=\text{constant}\]

This constant value is called the refractive index of the second medium with respect to the first.

REFRACTIVE INDEX

The refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium. It is represented by n.

\[n=\dfrac{\text{speed of light in vacuum/air}}{\text{speed of light in the medium}}=\dfrac{c}{v}\]

Both c and v are in m/s. The refractive index has no unit.

The relative refractive index of medium 2 with respect to medium 1 is

\[{}_1n_2=\dfrac{n_2}{n_1}=\dfrac{v_1}{v_2}\]

Both \(v_1\) and \(v_2\) are in m/s; \(n_1\) and \(n_2\) have no units.

\[{}_1n_2=\dfrac{1}{{}_2n_1}\]

Snell’s law of refraction: when light travels from medium 1 to medium 2, then

\[{}_1n_2=\dfrac{n_2}{n_1}=\dfrac{\sin i}{\sin r}\]

For an object seen normally through a medium:

\[{}_a n_w=\dfrac{\text{real depth}\,(x)}{\text{apparent depth}\,(a)}\]

Both x and a are in metre or in cm. The velocity of light in vacuum/air is \(c=3\times10^{8}\,\mathrm{m/s}\).

TWO REFRACTIONS THROUGH A RECTANGULAR GLASS SLAB

On passing through a rectangular glass slab, a ray of light suffers two refractions, one while going from air to glass and the other while going from glass to air. Light emerges from the rectangular slab in a direction parallel to that in which it entered the glass slab. However, the final emergent ray is slightly shifted sideways from the direction of the original incident ray by a distance x, called the lateral shift.

The perpendicular distance between the original path of the incident ray and the emergent ray coming out of the glass slab is called the lateral displacement of the emergent ray of light. Lateral displacement depends mainly on three factors: the angle of incidence, the thickness of the glass slab and the refractive index of the glass slab. Lateral displacement is directly proportional to (i) the angle of incidence, (ii) the thickness of the glass slab and (iii) the refractive index of the glass slab. The higher the values of these factors, the greater the lateral displacement. The angle which the emergent ray makes with the normal is called the angle of emergence.

CONDITION FOR NO REFRACTION

Refraction will not take place under the following two conditions:

1. When light is incident normally on a boundary. A ray of light travelling in medium 1 falls normally. Therefore the angle of incidence \(i=0^{\circ}\). According to Snell’s law,

\[\dfrac{\sin i}{\sin r}=\dfrac{n_2}{n_1}\;\Rightarrow\;\sin r=\dfrac{n_1}{n_2}\sin i=\dfrac{n_1}{n_2}\sin0^{\circ}=\dfrac{n_1}{n_2}\times0=0\;\Rightarrow\;r=0^{\circ}\]

Thus, there is no deviation in the ray at the boundary. Hence, no refraction occurs when light is incident normally on a boundary of two media.

2. When the refractive indices of two media are equal. When the refractive index of medium 1 is equal to the refractive index of medium 2, i.e. \(n_1=n_2\), then according to Snell’s law

\[\dfrac{\sin i}{\sin r}=\dfrac{n_2}{n_1}=1\;\Rightarrow\;\sin i=\sin r\;\Rightarrow\;i=r\]

Hence, no refraction occurs at the boundary that separates two media of equal refractive indices.

INTEXT QUESTIONS (PAGE 176)

A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer: The light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into water will bend towards the normal.

Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m/s.

Answer: The refractive index of a medium is given by \(n_m=\dfrac{c}{v}\).
\(c=3\times10^{8}\,\mathrm{m/s},\quad n_g=1.50\)
\(v=\dfrac{c}{n_g}=\dfrac{3\times10^{8}}{1.50}=2\times10^{8}\,\mathrm{m/s}\)

Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer: Highest optical density: diamond. Lowest optical density: air.
The optical density of a medium is directly related to the refractive index of that medium. A medium which has the highest refractive index will have the highest optical density and vice-versa. It can be observed from Table 10.3 that diamond and air respectively have the highest and the lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

Answer: \(n_m=\dfrac{c}{v}\;\Rightarrow\;v=\dfrac{c}{n_m}\;\Rightarrow\;v\propto\dfrac{1}{n_m}\)
It can be inferred from the relation that light will travel the slowest in the material which has the highest refractive index and the fastest in the material which has the lowest refractive index. The refractive indices of kerosene, turpentine and water are 1.44, 1.47 and 1.33 respectively. Therefore, light travels the fastest in water.

The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer: The refractive index of a medium is related to the speed of light in that medium by \(n_m=\dfrac{c}{v}\), where c is the speed of light in vacuum/air. The refractive index of diamond is 2.42. This means that the speed of light in diamond is reduced by a factor of 2.42 compared to its speed in air (vacuum).

NUMERICALS

Light travels through water with a speed of 2.25 × 10⁸ m/s. What is the refractive index of water?

Answer: \(n=\dfrac{c}{v}=\dfrac{3\times10^{8}}{2.25\times10^{8}}=1.33\)

Light travels from a rarer medium 1 to a denser medium 2. The angles of incidence and refraction are respectively 45° and 30°. Calculate (i) the refractive index of the second medium with respect to the first medium and (ii) the refractive index of medium 1 with respect to medium 2.

Answer: (i) \({}_1n_2=\dfrac{\sin i}{\sin r}=\dfrac{\sin45^{\circ}}{\sin30^{\circ}}=\dfrac{0.707}{0.5}=1.414\)
(ii) \({}_2n_1=\dfrac{1}{{}_1n_2}=\dfrac{1}{1.414}=0.707\)

A pond of depth 20 cm is filled with water of refractive index 4/3. Calculate the apparent depth of the tank when viewed normally.

Answer: \(\text{apparent depth}=\dfrac{\text{real depth}}{n}=\dfrac{20}{4/3}=15\,\mathrm{cm}\)

How much time will light take to cross a 2 mm thick glass pane if the refractive index of glass is 3/2?

Answer: \(v=\dfrac{c}{n}=\dfrac{3\times10^{8}}{3/2}=2\times10^{8}\,\mathrm{m/s}\)
\(t=\dfrac{d}{v}=\dfrac{2\times10^{-3}}{2\times10^{8}}=1\times10^{-11}\,\mathrm{s}\)

Calculate the speed of light in water of refractive index 4/3.

Answer: \(v=\dfrac{c}{n}=\dfrac{3\times10^{8}}{4/3}=2.25\times10^{8}\,\mathrm{m/s}\)

A ray of light passes from air to glass (n = 1.5) at an angle of 30°. Calculate the angle of refraction.

Answer: \(n=\dfrac{\sin i}{\sin r}\;\Rightarrow\;\sin r=\dfrac{\sin30^{\circ}}{1.5}=\dfrac{0.5}{1.5}=0.333\;\Rightarrow\;r=19.5^{\circ}\)

A ray of light is incident on a glass slab at an angle of 45°. If the refractive index of glass is 1.6, what is the angle of refraction?

Answer: \(\sin r=\dfrac{\sin45^{\circ}}{1.6}=\dfrac{0.707}{1.6}=0.442\;\Rightarrow\;r=26.2^{\circ}\)

The refractive index of diamond is 2.47 and that of glass is 1.51. How much faster does light travel in glass than in diamond?

Answer: \(v_{\text{glass}}=\dfrac{3\times10^{8}}{1.51}=1.99\times10^{8}\,\mathrm{m/s},\quad v_{\text{diamond}}=\dfrac{3\times10^{8}}{2.47}=1.21\times10^{8}\,\mathrm{m/s}\)
\(\text{Difference}=(1.987-1.215)\times10^{8}=7.7\times10^{7}\,\mathrm{m/s}\)
Light travels about \(7.7\times10^{7}\,\mathrm{m/s}\) faster in glass than in diamond (about 1.64 times as fast).

The refractive index of glycerine is 1.46. What is the speed of light in air if its speed in glycerine is 2.05 × 10⁸ m/s?

Answer: \(c=n\,v=1.46\times2.05\times10^{8}=2.99\times10^{8}\approx3\times10^{8}\,\mathrm{m/s}\)

The refractive index of glass is 1.6 and that of diamond is 2.4. Calculate (i) the refractive index of diamond with respect to glass and (ii) the refractive index of glass with respect to diamond.

Answer: (i) \({}_gn_d=\dfrac{2.4}{1.6}=1.5\)
(ii) \({}_dn_g=\dfrac{1.6}{2.4}=0.67\)

A ray of light is travelling from glass to air. The angle of incidence in glass is 30° and the angle of refraction in air is 60°. What is the refractive index of glass with respect to air?

Answer: \(n_g\sin30^{\circ}=n_a\sin60^{\circ}\;\Rightarrow\;n_g=\dfrac{1\times0.866}{0.5}=1.73\)

A ray of light is travelling from air to water. What is the angle of incidence in air, if the angle of refraction in water is 45°? Take the refractive index of water = 1.32.

Answer: \(\sin i=n\sin r=1.32\times\sin45^{\circ}=1.32\times0.707=0.933\;\Rightarrow\;i=69^{\circ}\)

A water tank appears to be 4 m deep when viewed from the top. If the refractive index of water is 4/3, what is the actual depth of the tank?

Answer: \(\text{real depth}=n\times\text{apparent depth}=\dfrac43\times4=5.33\,\mathrm{m}\)

What is the real depth of a swimming pool when its bottom appears to be raised by 1 m? The refractive index of water is 4/3.

Answer: \(\text{apparent depth}=\dfrac{x}{n}=\dfrac{3x}{4}\)
\(x-\dfrac{3x}{4}=1\;\Rightarrow\;\dfrac{x}{4}=1\;\Rightarrow\;x=4\,\mathrm{m}\)

A jar 15 cm long is filled with a transparent liquid. When viewed from the top, its bottom appears to be 12 cm below. What is the refractive index of the liquid?

Answer: \(n=\dfrac{\text{real depth}}{\text{apparent depth}}=\dfrac{15}{12}=1.25\)

SPHERICAL LENSES

A lens is any transparent material (e.g. glass) of an appropriate shape that can take parallel rays of incident light and either converge the rays to a point or diverge the rays from a point.

A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens.

Some lenses focus light rays to a single point. These lenses are called converging or convex lenses. Other lenses spread out the light rays so that it looks like they all come from the same point. These lenses are called diverging or concave lenses. Lenses change the direction of light rays by refraction. They are designed so that the image appears in a certain place or as a certain size. Lenses are used in eyeglasses, cameras, microscopes and telescopes.

CONVEX LENS

A lens may have two spherical surfaces bulging outwards. Such a lens is called a double convex lens. It is simply called a convex lens. It is thicker at the middle as compared to the edges. A convex lens converges light rays. Hence it is called a converging lens.

CONCAVE LENS

A double concave lens is bounded by two spherical surfaces, curved inwards. It is thicker at the edges than at the middle. Such lenses diverge light rays and are called diverging lenses. A double concave lens is simply called a concave lens.

TERMS RELATED TO SPHERICAL LENSES

Principal axis: The principal axis is the line which runs horizontally straight through the optical centre of the lens. It is also sometimes called the optic axis. In other words, an imaginary straight line passing through the two centres of curvature of a lens is called its principal axis.

Optical centre: The optical centre (O) of a convex lens is usually the centre point of the lens. The direction of all light rays which pass through the optical centre remains unchanged.

Centre of curvature: A lens has two spherical surfaces. Each of these surfaces forms a part of a sphere. The centres of these spheres are called centres of curvature of the lens. The centre of curvature of a lens is usually represented by the letter C. Since there are two centres of curvature, we may represent them as \(C_1\) and \(C_2\).

Aperture: The effective diameter of the circular outline of a spherical lens is called its aperture. Lenses whose aperture is much less than the radius of curvature are called thin lenses with small aperture.

Focus: The focus or focal point of the lens is the position on the principal axis where all light rays that run parallel to the principal axis through the lens converge (come together) at a point. Since light can pass through the lens either from right to left or left to right, there is a focal point on each side of the lens (\(F_1\) and \(F_2\)), at the same distance from the optical centre in each direction. (Note: the plural form of the word focus is foci.)

Focal length: The focal length (f) is the distance between the optical centre and the focal point.

RULES FOR OBTAINING IMAGES FORMED BY SPHERICAL LENSES

The intersection of at least two refracted rays gives the position of the image of the point object. Any two of the following rays can be considered for locating the image.

1. A ray of light from the object, parallel to the principal axis, after refraction from a convex lens, passes through the principal focus on the other side of the lens. In the case of a concave lens, the ray appears to diverge from the principal focus located on the same side of the lens.
2. A ray of light passing through a principal focus, after refraction from a convex lens, will emerge parallel to the principal axis. A ray of light appearing to meet at the principal focus of a concave lens, after refraction, will emerge parallel to the principal axis.
3. A ray of light passing through the optical centre of a lens will emerge without any deviation.

FORMATION OF DIFFERENT TYPES OF IMAGES BY A CONVEX LENS

The type of image formed by a convex lens depends on the position of the object in front of the lens. There are six positions of the object:

Case 1: Object is in between the optical centre (O) and the focus (F₁)

1. on the same side of the lens as the object (on the left of the lens)
2. virtual and erect, and
3. larger than the object (enlarged or magnified).

Case 2: Object is at the focus (F₁)

1. at infinity
2. real and inverted, and
3. highly enlarged.

Case 3: Object is in between F₁ and 2F₁

1. beyond 2F₂
2. real and inverted, and
3. larger than the object (or magnified).

Case 4: Object is at 2F₁

1. at 2F₂ on the other side of the lens
2. real and inverted, and
3. of the same size as the object.

Case 5: Object is beyond 2F₁

1. between F₂ and 2F₂ on the other side of the lens
2. real and inverted, and
3. smaller than the object (or diminished).

Case 6: Object is at infinity

1. at the focus F₂
2. real and inverted, and
3. much smaller than the object (or highly diminished or point-sized).

FORMATION OF DIFFERENT TYPES OF IMAGES BY A CONCAVE LENS

The type of image formed by a concave lens depends on the position of the object in front of the lens. There are two positions of the object:

Case 1: Object is at infinity

1. at the focus F₁
2. virtual and erect, and
3. much smaller than the object (or highly diminished or point-sized).

Case 2: Object is in between the optical centre (O) and infinity

1. between the optical centre (O) and the focus F₁
2. virtual and erect, and
3. smaller than the object (or diminished).

SIGN CONVENTION FOR SPHERICAL LENSES

While using the lens formula we must make use of the proper sign convention while taking the values of object distance (u), image distance (v), focal length (f), object height (h) and image height (h′). The sign conventions are as follows:

1. All distances are measured from the optical centre of the lens.
2. Distances measured in the same direction as the incident light are taken positive.
3. Distances measured in the direction opposite to the direction of incident light are taken negative.
4. Heights measured upwards and perpendicular to the principal axis are taken positive.
5. Heights measured downwards and perpendicular to the principal axis are taken negative.

Consequences of the new Cartesian sign convention:

• The focal length of a convex lens is positive and that of a concave lens is negative.
• The object distance u is always negative.
• The distance of a real image is positive and that of a virtual image is negative.
• The object height h is always positive. The height h′ of a virtual erect image is positive and that of a real inverted image is negative.
• The linear magnification m = h′/h is positive for a virtual image and negative for a real image.

LENS FORMULA

The lens formula gives the relationship between the object distance (u), the image distance (v) and the focal length (f):

\[\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\]

where u is the distance of the object from the optical centre (O), v is the distance of the image from the optical centre (O) and f is the distance of the principal focus from the optical centre (O).

MAGNIFICATION

The magnification produced by a lens, similar to that for spherical mirrors, is defined as the ratio of the height of the image to the height of the object. It is represented by the letter m. If h is the height of the object and h′ is the height of the image given by a lens, then the magnification produced by the lens is given by

\[m=\dfrac{\text{height of the image}}{\text{height of the object}}=\dfrac{h\prime}{h}\]

Magnification produced by a lens is also related to the object distance u and the image distance v. This relationship is given by

\[m=\dfrac{h\prime}{h}=\dfrac{v}{u}\]

Points to be remembered

• If the magnification m has a positive value, the image is virtual and erect. If the magnification m has a negative value, the image is real and inverted.
• A convex lens can form virtual images as well as real images; therefore, the magnification produced by a convex lens can be either positive or negative.
• A convex lens can form images which are smaller than the object, equal to the object or bigger than the object; therefore the magnification m produced by a convex lens can be less than 1, equal to 1 or more than 1.
• A concave lens, however, forms only virtual images, so the magnification produced by a concave lens is always positive.
• A concave lens forms images which are always smaller than the object, so the magnification m produced by a concave lens is always less than 1.

NUMERICALS BASED ON CONVEX LENS

A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall?

Answer: \(f=+10\,\mathrm{cm},\; v=+12\,\mathrm{cm}\)
\(\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=\dfrac{1}{12}-\dfrac{1}{10}=-\dfrac{1}{60}\;\Rightarrow\;u=-60\,\mathrm{cm}\)
The object should be placed 60 cm from the lens.

If an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8 cm, find the position, nature and height of the image.

Answer: \(u=-12\,\mathrm{cm},\; f=+8\,\mathrm{cm},\; h=7\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{8}-\dfrac{1}{12}=\dfrac{1}{24}\;\Rightarrow\;v=+24\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{24}{-12}=-2\;\Rightarrow\;h\prime=-14\,\mathrm{cm}\)
The image is formed 24 cm from the lens on the other side; it is real, inverted and magnified, 14 cm in height.

An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.

Answer: \(\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=-\dfrac{1}{20}\;\Rightarrow\;v=-20\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{-20}{-10}=+2\;\Rightarrow\;h\prime=+8\,\mathrm{cm}\)
The image is formed 20 cm from the lens on the same side as the object; it is virtual, erect and magnified, 8 cm in height.

A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.

Answer: \(f=+5\,\mathrm{cm},\; v=-25\,\mathrm{cm}\)
\(\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{25}-\dfrac{1}{5}=-\dfrac{6}{25}\;\Rightarrow\;u=-4.17\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{-25}{-4.17}=+6\)

Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.

Answer: \(\dfrac{1}{v}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\;\Rightarrow\;v=+15\,\mathrm{cm}\)
\(m=\dfrac{15}{-10}=-1.5\;\Rightarrow\;h\prime=-7.5\,\mathrm{cm}\)
The image is 15 cm from the lens on the other side; it is real, inverted and magnified (7.5 cm high).

Calculate the focal length of a convex lens which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it.

Answer: \(u=-20\,\mathrm{cm},\; v=-50\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{50}+\dfrac{1}{20}=\dfrac{3}{100}\;\Rightarrow\;f=+33.3\,\mathrm{cm}\)

An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. What is the nature and position of the image?

Answer: \(\dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{100}=\dfrac{3}{200}\;\Rightarrow\;v=+66.7\,\mathrm{cm}\)
\(m=\dfrac{66.7}{-100}=-0.67\)
The image is real and inverted, diminished, 66.7 cm from the lens on the other side.

A convex lens produces an inverted image magnified three times of an object at a distance of 15 cm from it. Calculate the focal length of the lens.

Answer: \(m=-3=\dfrac{v}{u}\;\Rightarrow\;v=-3u=+45\,\mathrm{cm}\;(u=-15\,\mathrm{cm})\)
\(\dfrac{1}{f}=\dfrac{1}{45}+\dfrac{1}{15}=\dfrac{4}{45}\;\Rightarrow\;f=11.25\,\mathrm{cm}\)

An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens. What is the magnification of the image? What is the focal length of the lens? Also draw the ray diagram to show the formation of the image.

Answer: \(u=-4\,\mathrm{cm},\; v=+12\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{12}{-4}=-3\)
\(\dfrac{1}{f}=\dfrac{1}{12}+\dfrac{1}{4}=\dfrac{1}{3}\;\Rightarrow\;f=3\,\mathrm{cm}\)
The object lies between F and 2F, so the image is real, inverted, magnified and beyond 2F on the other side.

A lens of focal length 20 cm is used to produce a ten times magnified image of a film slide on a screen. How far must the slide be placed from the lens?

Answer: \(m=-10\;\Rightarrow\;v=-10u\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{10u}-\dfrac{1}{u}=-\dfrac{11}{10u}\;\Rightarrow\;u=-\dfrac{11f}{10}=-22\,\mathrm{cm}\)
The slide must be placed 22 cm from the lens (the image is on the screen at 220 cm).

Determine how far an object must be placed in front of a converging lens of focal length 10 cm in order to produce an erect image of linear magnification 4.

Answer: \(m=+4\;\Rightarrow\;v=4u\)
\(\dfrac{1}{f}=\dfrac{1}{4u}-\dfrac{1}{u}=-\dfrac{3}{4u}\;\Rightarrow\;u=-\dfrac{3f}{4}=-7.5\,\mathrm{cm}\)
The object must be 7.5 cm from the lens.

A convex lens of focal length 6 cm is held 4 cm from a newspaper, which has print 0.5 cm high. By calculation, determine the size and nature of the image produced.

Answer: \(\dfrac{1}{v}=\dfrac{1}{6}-\dfrac{1}{4}=-\dfrac{1}{12}\;\Rightarrow\;v=-12\,\mathrm{cm}\)
\(m=\dfrac{-12}{-4}=+3\;\Rightarrow\;h\prime=1.5\,\mathrm{cm}\)
The image is virtual, erect and magnified, 1.5 cm high, on the same side of the lens as the newspaper.

A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Find the position, nature and size of the image.

Answer: \(\dfrac{1}{v}=\dfrac{1}{0.10}-\dfrac{1}{0.08}=10-12.5=-2.5\;\Rightarrow\;v=-0.40\,\mathrm{m}\)
\(m=\dfrac{-0.40}{-0.08}=+5\;\Rightarrow\;h\prime=25\,\mathrm{mm}\)
The image is 40 cm from the lens on the same side as the object; it is virtual, erect and magnified, 25 mm high.

An erect image 2 cm high is formed 12 cm from a lens, the object being 0.5 cm high. Find the focal length of the lens.

Answer: \(m=\dfrac{2}{0.5}=+4=\dfrac{v}{u}\;\Rightarrow\;u=\dfrac{v}{4}=\dfrac{-12}{4}=-3\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{12}+\dfrac{1}{3}=\dfrac{1}{4}\;\Rightarrow\;f=+4\,\mathrm{cm}\)
The lens is convex, of focal length 4 cm (an erect image of an object is virtual, so it is on the same side as the object).

The filament of a lamp is 80 cm from a screen and a converging lens forms an image of it on the screen, magnified three times. Find the distance of the lens from the filament and the focal length of the lens.

Answer: The image is real, so \(m=-3\) and \(v=3|u|\). Since the filament and the screen are 80 cm apart, \(|u|+v=80\,\mathrm{cm}\).
\(|u|+3|u|=80\;\Rightarrow\;|u|=20\,\mathrm{cm},\quad v=60\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{60}+\dfrac{1}{20}=\dfrac{4}{60}\;\Rightarrow\;f=15\,\mathrm{cm}\)
The lens is 20 cm from the filament and its focal length is 15 cm.

An object 2 cm tall is placed on the axis of a convex lens of focal length 5 cm at a distance of 10 cm from the optical centre of the lens. Find the nature, position and size of the image formed. Which case of image formation by convex lenses is illustrated by this example?

Answer: \(\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{10}=\dfrac{1}{10}\;\Rightarrow\;v=+10\,\mathrm{cm}\)
\(m=\dfrac{10}{-10}=-1\;\Rightarrow\;h\prime=-2\,\mathrm{cm}\)
The image is real, inverted and of the same size (2 cm), formed 10 cm from the lens on the other side. This illustrates the case in which the object is placed at 2F₁ (Case 4).

A converging lens of focal length 5 cm is placed at a distance of 20 cm from a screen. How far from the lens should an object be placed so as to form its real image on the screen?

Answer: \(v=+20\,\mathrm{cm},\; f=+5\,\mathrm{cm}\)
\(\dfrac{1}{u}=\dfrac{1}{20}-\dfrac{1}{5}=-\dfrac{3}{20}\;\Rightarrow\;u=-6.67\,\mathrm{cm}\)
The object should be placed 6.67 cm from the lens.

An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.

Answer: \(\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{25}=\dfrac{3}{50}\;\Rightarrow\;v=+16.7\,\mathrm{cm}\)
\(m=\dfrac{16.7}{-25}=-0.67\;\Rightarrow\;h\prime=-3.3\,\mathrm{cm}\)
The image is real, inverted and diminished (3.3 cm), formed 16.7 cm from the lens on the other side.

At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side? What will be the magnification produced in this case?

Answer: \(v=+24\,\mathrm{cm},\; f=+18\,\mathrm{cm}\)
\(\dfrac{1}{u}=\dfrac{1}{24}-\dfrac{1}{18}=-\dfrac{1}{72}\;\Rightarrow\;u=-72\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{24}{-72}=-\dfrac13\)
The object should be 72 cm from the lens; the image is real, inverted and diminished.

The magnification produced by a spherical lens is +2.5. What is the nature of the image and the lens?

Answer: Positive magnification means the image is virtual and erect; since it is magnified (2.5 times), the lens is a convex lens (a concave lens always gives a diminished image).

What is the nature of the image formed by a convex lens if the magnification produced is +3?

Answer: The image is virtual, erect and magnified (three times the size of the object); the object is between the optical centre and the focus.

What is the nature of the image formed by a convex lens if the magnification produced is −0.5?

Answer: The negative sign shows that the image is real and inverted; the magnitude 0.5 shows that it is diminished (half the size of the object).

What is the position of the image when an object is placed at a distance of 10 cm from a convex lens of focal length 10 cm?

Answer: \(\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{10}-\dfrac{1}{10}=0\;\Rightarrow\;v=\infty\)
The object is at the focus, so the refracted rays are parallel and the image is formed at infinity.

Describe the nature of the image formed when an object is placed at a distance of 30 cm from a convex lens of focal length 15 cm.

Answer: \(\dfrac{1}{v}=\dfrac{1}{15}-\dfrac{1}{30}=\dfrac{1}{30}\;\Rightarrow\;v=+30\,\mathrm{cm}\)
\(m=\dfrac{30}{-30}=-1\)
The object is at 2F; the image is real, inverted and of the same size as the object, formed at 30 cm on the other side of the lens.

At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced?

Answer: A convex lens gives a real image of the same size only when the object is at 2F, where \(m=-1\):
\(u=-2f=-24\,\mathrm{cm}\)
The object must be placed 24 cm from the lens.

NUMERICALS BASED ON CONCAVE LENS

A concave lens produces an image 20 cm from the lens of an object placed 30 cm from the lens. Calculate the focal length of the lens.

Answer: \(u=-30\,\mathrm{cm},\; v=-20\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{20}+\dfrac{1}{30}=-\dfrac{1}{60}\;\Rightarrow\;f=-60\,\mathrm{cm}\)

The magnification of a spherical lens is +0.5. What is the nature of the lens and image?

Answer: The image is virtual and erect, and it is diminished (half the size). Only a concave lens always forms a virtual, erect, diminished image, so the lens is a concave lens.

If an object is placed at a distance of 50 cm from a concave lens of focal length 20 cm, find the position, nature and height of the image.

Answer: \(u=-50\,\mathrm{cm},\; f=-20\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=-\dfrac{1}{20}-\dfrac{1}{50}=-\dfrac{7}{100}\;\Rightarrow\;v=-14.3\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{-14.3}{-50}=+0.29\)
The image is virtual, erect and diminished, 14.3 cm from the lens on the same side as the object. Its height is 0.29 times the height of the object (the object height is not given).

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.

Answer: \(\dfrac{1}{v}=-\dfrac{1}{12}-\dfrac{1}{4}=-\dfrac{1}{3}\;\Rightarrow\;v=-3\,\mathrm{cm}\)
\(m=\dfrac{-3}{-4}=+0.75\)
The image is virtual, erect and diminished, 3 cm from the lens on the same side as the object.

An object placed at a distance of 50 cm from a concave lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of the image. Calculate the focal length of the lens and the magnification produced.

Answer: \(u=-50\,\mathrm{cm},\; v=-10\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{10}+\dfrac{1}{50}=-\dfrac{4}{50}\;\Rightarrow\;f=-12.5\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{-10}{-50}=+0.2\)

A 50 cm tall object is at a very large distance from a diverging lens. A virtual, erect and diminished image of the object is formed at a distance of 20 cm in front of the lens. How much is the focal length of the lens?

Answer: For an object at a very large distance, the image is formed at the principal focus. So the focal length is 20 cm, i.e. \(f=-20\,\mathrm{cm}\) (diverging lens).

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer: \(f=-15\,\mathrm{cm},\; v=-10\,\mathrm{cm}\)
\(\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{10}+\dfrac{1}{15}=-\dfrac{1}{30}\;\Rightarrow\;u=-30\,\mathrm{cm}\)
The object is 30 cm in front of the lens.

An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or diverging? Give reasons for your answer.

Answer: \(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-\dfrac{1}{20}+\dfrac{1}{60}=-\dfrac{1}{30}\;\Rightarrow\;f=-30\,\mathrm{cm}\)
The focal length is negative and the image is virtual and smaller than the object, so the lens is a diverging (concave) lens.

A concave lens of 20 cm focal length forms an image 15 cm from the lens. Compute the object distance.

Answer: \(\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{15}+\dfrac{1}{20}=-\dfrac{1}{60}\;\Rightarrow\;u=-60\,\mathrm{cm}\)

A concave lens has focal length 15 cm. At what distance should the object be placed from the lens so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.

Answer: \(\dfrac{1}{u}=-\dfrac{1}{10}+\dfrac{1}{15}=-\dfrac{1}{30}\;\Rightarrow\;u=-30\,\mathrm{cm}\)
\(m=\dfrac{v}{u}=\dfrac{-10}{-30}=+\dfrac13\)

Calculate the image distance for an object of height 12 mm at a distance of 0.20 m from a concave lens of focal length 0.30 m and state the nature and size of the image.

Answer: \(\dfrac{1}{v}=-\dfrac{1}{0.30}-\dfrac{1}{0.20}=-\dfrac{25}{3}\;\Rightarrow\;v=-0.12\,\mathrm{m}=-12\,\mathrm{cm}\)
\(m=\dfrac{-0.12}{-0.20}=+0.6\;\Rightarrow\;h\prime=7.2\,\mathrm{mm}\)
The image is virtual and erect, 12 cm from the lens on the object side, and diminished (7.2 mm).

A concave lens has a focal length of 20 cm. At what distance from the lens should a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

Answer: \(\dfrac{1}{u}=-\dfrac{1}{15}+\dfrac{1}{20}=-\dfrac{1}{60}\;\Rightarrow\;u=-60\,\mathrm{cm}\)
\(m=\dfrac{-15}{-60}=+0.25\;\Rightarrow\;h\prime=1.25\,\mathrm{cm}\)

An object is placed 20 cm from (a) a converging lens and (b) a diverging lens of focal length 15 cm. Calculate the image position and magnification in each case.

Answer: (a) \(\dfrac{1}{v}=\dfrac{1}{15}-\dfrac{1}{20}=\dfrac{1}{60}\;\Rightarrow\;v=+60\,\mathrm{cm},\quad m=\dfrac{60}{-20}=-3\)
(b) \(\dfrac{1}{v}=-\dfrac{1}{15}-\dfrac{1}{20}=-\dfrac{7}{60}\;\Rightarrow\;v=-8.57\,\mathrm{cm},\quad m=\dfrac{-8.57}{-20}=+0.43\)

A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.

Answer: \(\dfrac{1}{v}=-\dfrac{1}{15}-\dfrac{1}{40}=-\dfrac{11}{120}\;\Rightarrow\;v=-10.9\,\mathrm{cm}\)
\(m=\dfrac{-10.9}{-40}=+0.27\;\Rightarrow\;h\prime=0.55\,\mathrm{cm}\)
The image is virtual and erect, 10.9 cm from the lens on the object side, and about 0.55 cm tall.

Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from (a) a diverging lens of focal length 40 cm and (b) a converging lens of focal length 40 cm.

Answer: (a) \(\dfrac{1}{v}=-\dfrac{1}{40}-\dfrac{1}{20}=-\dfrac{3}{40}\;\Rightarrow\;v=-13.3\,\mathrm{cm},\quad m=\dfrac{-13.3}{-20}=+0.67\;\Rightarrow\;h\prime=1.33\,\mathrm{cm}\)
(b) \(\dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{20}=-\dfrac{1}{40}\;\Rightarrow\;v=-40\,\mathrm{cm},\quad m=\dfrac{-40}{-20}=+2\;\Rightarrow\;h\prime=4\,\mathrm{cm}\)

The magnification produced by a spherical lens is +0.75. What is the nature of the image and the lens?

Answer: The image is virtual and erect, and diminished (0.75 times). The lens is a concave lens.

The magnification produced by a spherical lens and a spherical mirror is +0.8. What is the nature of the lens and the mirror?

Answer: Positive magnification below 1 means a virtual, erect, diminished image. The lens is a concave lens and the mirror is a convex mirror (a concave mirror gives a virtual image only when it is magnified).

The magnification produced by a spherical lens and a spherical mirror is +2.0. What is the nature of the lens and the mirror?

Answer: Positive magnification greater than 1 means a virtual, erect, magnified image. The lens is a convex lens and the mirror is a concave mirror.

The lens A produces a magnification of −0.6, whereas lens B produces a magnification of +0.6. What is the nature of lens A and B?

Answer: Lens A: the negative magnification shows a real, inverted, diminished image, so A is a convex lens.
Lens B: the positive magnification (less than 1) shows a virtual, erect, diminished image, so B is a concave lens.

An object is 2 m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?

Answer: \(u=-2\,\mathrm{m},\quad m=+\dfrac14=\dfrac{v}{u}\;\Rightarrow\;v=-0.5\,\mathrm{m}\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=-2+0.5=-1.5\;\Rightarrow\;f=-0.67\,\mathrm{m}=-67\,\mathrm{cm}\)
The focal length is negative, so the lens is a concave (diverging) lens.

POWER OF A LENS

The power of a lens is defined as the reciprocal of its focal length. It is represented by the letter P. The power P of a lens of focal length f is given by

\[P=\dfrac{1}{f}\]

The SI unit of power of a lens is the dioptre. It is denoted by the letter D. If f is expressed in metres, then power is expressed in dioptres. Thus, 1 dioptre is the power of a lens whose focal length is 1 metre. \(1\,\mathrm{D}=1\,\mathrm{m}^{-1}\). The power of a convex lens is positive and that of a concave lens is negative.

Many optical instruments consist of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses placed in contact is given by the algebraic sum of the individual powers \(P_1,\,P_2,\,P_3,\,\ldots\) as

\[P=P_1+P_2+P_3+\cdots\]

NUMERICALS ON POWER OF A LENS

A concave lens produces an image 20 cm from the lens of an object placed 30 cm from the lens. Calculate the power of the lens.

Answer: From the earlier problem \(f=-60\,\mathrm{cm}=-0.6\,\mathrm{m}\).
\(P=\dfrac{1}{f}=\dfrac{1}{-0.6}=-1.67\,\mathrm{D}\)

A convex lens is of focal length 10 cm. What is its power?

Answer: \(P=\dfrac{1}{0.10}=+10\,\mathrm{D}\)

A person having a myopic eye uses a concave lens of focal length 50 cm. What is the power of the lens?

Answer: \(P=\dfrac{1}{-0.50}=-2\,\mathrm{D}\)

A thin lens has a focal length of −25 cm. What is the power of the lens and what is its nature?

Answer: \(P=\dfrac{1}{-0.25}=-4\,\mathrm{D}\)
The power is negative, so it is a concave (diverging) lens.

A lens has a power of −2.5 D. What is the focal length and nature of the lens?

Answer: \(f=\dfrac{1}{P}=\dfrac{1}{-2.5}=-0.4\,\mathrm{m}=-40\,\mathrm{cm}\)
The lens is concave (diverging).

Find the power of a concave lens of focal length 2 m.

Answer: \(f=-2\,\mathrm{m}\;\Rightarrow\;P=\dfrac{1}{-2}=-0.5\,\mathrm{D}\)

A convex lens forms a real and inverted image of a needle at a distance of 50 cm from the lens. If the image is of the same size as the needle, where is the needle placed in front of the lens? Also, find the power of the lens.

Answer: The image is real and of the same size, so the needle is at 2F; hence the needle is 50 cm in front of the lens. \(u=-50\,\mathrm{cm},\; v=+50\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{50}+\dfrac{1}{50}=\dfrac{1}{25}\;\Rightarrow\;f=25\,\mathrm{cm}=0.25\,\mathrm{m}\)
\(P=\dfrac{1}{0.25}=+4\,\mathrm{D}\)

Two thin lenses of power +3.5 D and −2.5 D are placed in contact. Find the power and focal length of the lens combination.

Answer: \(P=3.5-2.5=+1\,\mathrm{D}\)
\(f=\dfrac{1}{P}=1\,\mathrm{m}=100\,\mathrm{cm}\)

A doctor has prescribed a corrective lens of power −1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer: \(f=\dfrac{1}{-1.5}=-0.67\,\mathrm{m}=-67\,\mathrm{cm}\)
A concave lens has a negative focal length; hence the prescribed lens is a diverging lens.

A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed in contact with each other. What is the power of this combination? Also, calculate the focal length of the combination.

Answer: \(P_1=\dfrac{1}{-0.25}=-4\,\mathrm{D},\quad P_2=\dfrac{1}{0.20}=+5\,\mathrm{D}\)
\(P=P_1+P_2=+1\,\mathrm{D},\quad f=\dfrac1P=1\,\mathrm{m}=100\,\mathrm{cm}\)

A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 10 cm. What is the focal length and power of the combination?

Answer: \(P_1=+5\,\mathrm{D},\quad P_2=-10\,\mathrm{D}\;\Rightarrow\;P=-5\,\mathrm{D}\)
\(f=\dfrac{1}{-5}=-0.2\,\mathrm{m}=-20\,\mathrm{cm}\)
The combination is a diverging lens.

An object is placed at a distance of 50 cm from a concave lens of focal length 30 cm. Find the nature and position of the image.

Answer: \(\dfrac{1}{v}=-\dfrac{1}{30}-\dfrac{1}{50}=-\dfrac{8}{150}\;\Rightarrow\;v=-18.75\,\mathrm{cm}\)
The image is virtual, erect and diminished, 18.75 cm from the lens on the same side as the object.

An object of height 2 cm is placed at a distance of 15 cm in front of a concave lens of power −10 D. Find the size of the image.

Answer: \(f=\dfrac{1}{-10}=-0.1\,\mathrm{m}=-10\,\mathrm{cm}\)
\(\dfrac{1}{v}=-\dfrac{1}{10}-\dfrac{1}{15}=-\dfrac{1}{6}\;\Rightarrow\;v=-6\,\mathrm{cm}\)
\(m=\dfrac{-6}{-15}=0.4\;\Rightarrow\;h\prime=0.8\,\mathrm{cm}\)
The image is virtual, erect and 0.8 cm in size.

A convergent lens of power 8 D is combined with a divergent lens of power −10 D. Calculate the focal length of the combination.

Answer: \(P=8-10=-2\,\mathrm{D}\)
\(f=\dfrac{1}{-2}=-0.5\,\mathrm{m}=-50\,\mathrm{cm}\)

A concave lens is kept in contact with a convex lens of focal length 20 cm. The combination works as a converging lens of focal length 100 cm. Calculate the power of the concave lens.

Answer: \(P=\dfrac{1}{1}=+1\,\mathrm{D},\quad P_{\text{convex}}=\dfrac{1}{0.2}=+5\,\mathrm{D}\)
\(P_{\text{concave}}=P-P_{\text{convex}}=1-5=-4\,\mathrm{D}\)

Find the focal length and nature of the lens which should be placed in contact with a lens of focal length 10 cm so that the power of the combination becomes 5 D.

Answer: \(P_1=\dfrac{1}{0.10}=+10\,\mathrm{D},\quad P_2=P-P_1=5-10=-5\,\mathrm{D}\)
\(f_2=\dfrac{1}{-5}=-0.2\,\mathrm{m}=-20\,\mathrm{cm}\)
The lens to be added is a concave lens of focal length 20 cm.

A convex lens of power 3 D is held in contact with a concave lens of power −1 D. A parallel beam of light is made to fall on the combination. At what distance from the combination will the beam get focussed?

Answer: \(P=3-1=+2\,\mathrm{D},\quad f=\dfrac{1}{2}=0.5\,\mathrm{m}=50\,\mathrm{cm}\)
A parallel beam is focussed at the focal length, i.e. 50 cm from the combination.

A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with one another. (a) What is the power of the combination? (b) What is the focal length of the combination? (c) Is this combination converging or diverging?

Answer: \(P_1=\dfrac{1}{0.25}=+4\,\mathrm{D},\quad P_2=\dfrac{1}{-0.10}=-10\,\mathrm{D}\)
(a) \(P=4-10=-6\,\mathrm{D}\)
(b) \(f=\dfrac{1}{-6}=-0.167\,\mathrm{m}=-16.7\,\mathrm{cm}\)
(c) The power is negative, so the combination is diverging.

The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X is 15 cm, then (a) calculate the focal length of lens Y; (b) state the nature of lens Y.

Answer: \(P_X=\dfrac{1}{0.15}=+6.67\,\mathrm{D},\quad P_Y=5-6.67=-1.67\,\mathrm{D}\)
(a) \(f_Y=\dfrac{1}{-1.67}=-0.6\,\mathrm{m}=-60\,\mathrm{cm}\)
(b) Lens Y is a concave lens.

Two lenses A and B have focal lengths of +20 cm and −10 cm respectively. (a) What is the nature of lens A and lens B? (b) What is the power of lens A and lens B? What is the power of the combination if lenses A and B are held close together?

Answer: (a) A is a convex lens and B is a concave lens.
(b) \(P_A=\dfrac{1}{0.20}=+5\,\mathrm{D},\quad P_B=\dfrac{1}{-0.10}=-10\,\mathrm{D}\)
Power of the combination: \(P=5-10=-5\,\mathrm{D}\)

INTEXT QUESTIONS (PAGE 184)

Define 1 dioptre of power of a lens.

Answer: The power of a lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length f in metres, then \(P=\dfrac{1}{f\,(\text{in metres})}\). The SI unit of power of a lens is the dioptre, denoted by D. 1 dioptre is defined as the power of a lens of focal length 1 metre. Hence, \(1\,\mathrm{D}=1\,\mathrm{m}^{-1}\).

A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Answer: When an object is placed at the centre of curvature, 2F₁, of a convex lens, its image is formed at the centre of curvature, 2F₂, on the other side of the lens. The image formed is inverted and of the same size as the object.

Answer: The image is formed at a distance of 50 cm from the convex lens, so the needle is placed 50 cm in front of the lens.
\(u=-50\,\mathrm{cm},\; v=+50\,\mathrm{cm}\)
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{50}+\dfrac{1}{50}=\dfrac{1}{25}\;\Rightarrow\;f=25\,\mathrm{cm}=0.25\,\mathrm{m}\)
\(P=\dfrac{1}{f\,(\text{in metres})}=\dfrac{1}{0.25}=+4\,\mathrm{D}\)
Hence, the power of the given lens is +4 D.

Find the power of a concave lens of focal length 2 m.

Answer: Focal length of the concave lens, \(f=-2\,\mathrm{m}\).
\(P=\dfrac{1}{f\,(\text{in metres})}=\dfrac{1}{-2}=-0.5\,\mathrm{D}\)
The negative sign arises due to the divergent nature of the concave lens. Hence, the power of the given concave lens is −0.5 D.

EXERCISE QUESTIONS (PAGES 185 AND 186)

Which one of the following materials cannot be used to make a lens?

1. Water
2. Glass
3. Plastic
4. Clay

Answer: (d) A lens allows light to pass through it. Since clay does not show such a property, it cannot be used to make a lens.

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the position of the object be?

1. Between the principal focus and the centre of curvature
2. At the centre of curvature
3. Beyond the centre of curvature
4. Between the pole of the mirror and its principal focus

Answer: (d) When an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect and larger than the object.

Where should an object be placed in front of a convex lens to get a real image of the size of the object?

1. At the principal focus of the lens
2. At twice the focal length
3. At infinity
4. Between the optical centre of the lens and its principal focus

Answer: (b) When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens. The image formed is real, inverted and of the same size as the object.

A spherical mirror and a thin spherical lens have each a focal length of −15 cm. The mirror and the lens are likely to be

1. both concave
2. both convex
3. the mirror is concave and the lens is convex
4. the mirror is convex, but the lens is concave

Answer: (a) By convention, the focal length of a concave mirror and of a concave lens is taken as negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

1. plane
2. concave
3. convex
4. either plane or convex

Answer: (d) A convex mirror always gives a virtual and erect image of smaller size of the object placed in front of it. Similarly, a plane mirror will always give a virtual and erect image of the same size as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

1. A convex lens of focal length 50 cm
2. A concave lens of focal length 50 cm
3. A convex lens of focal length 5 cm
4. A concave lens of focal length 5 cm

Answer: (c) A convex lens gives a magnified image of an object when it is placed between the optical centre and the focus (within the focal length). Also, magnification is more for convex lenses having shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.

We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer: Range of object distance = 0 cm to 15 cm.
A concave mirror gives an erect image when an object is placed between its pole (P) and the principal focus (F). Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the object must be placed anywhere between the pole and the focus. The image formed will be virtual, erect and magnified, as shown in the figure.

Name the type of mirror used in the following situations. (a) Headlights of a car. (b) Side/rear-view mirror of a vehicle. (c) Solar furnace. Support your answer with reason.

Answer: (a) Concave (b) Convex (c) Concave
Explanation:
(a) A concave mirror is used in the headlights of a car. This is because concave mirrors can produce a powerful parallel beam of light when the light source is placed at their principal focus.
(b) A convex mirror is used as the side/rear-view mirror of a vehicle. Convex mirrors give a virtual, erect and diminished image of the objects placed in front of them. Because of this, they have a wide field of view. It enables the driver to see most of the traffic behind him/her.
(c) Concave mirrors are convergent mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point known as the principal focus. Hence, they can be used to produce a large amount of heat at that point.

One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer: The convex lens will form a complete image of an object even if one half of it is covered with black paper. It can be understood by the following two cases.

Case I: When the upper half of the lens is covered

In this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet on the other side of the lens to form the image of the given object, as shown in the figure.

Case II: When the lower half of the lens is covered

In this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet on the other side of the lens to form the image of the given object, as shown in the figure.

In both cases the image is complete; only its brightness (intensity) decreases, because fewer rays reach the image.

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer: \(u=-25\,\mathrm{cm},\quad h=5\,\mathrm{cm},\quad f=+10\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{10}-\dfrac{1}{25}=\dfrac{5-2}{50}=\dfrac{3}{50}\;\Rightarrow\;v=\dfrac{50}{3}=16.67\,\mathrm{cm}\)
The positive value of v shows that the image is formed on the other side of the lens.
\(m=\dfrac{v}{u}=\dfrac{16.67}{-25}=-0.67\)
The negative sign shows that the image is real and formed behind the lens.
\(m=\dfrac{h\prime}{h}\;\Rightarrow\;h\prime=-0.67\times5=-3.3\,\mathrm{cm}\)
The negative value of the image height indicates that the image formed is inverted. The position, size and nature of the image are shown in the ray diagram.

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer: Focal length of the concave lens, \(f=-15\,\mathrm{cm}\); image distance, \(v=-10\,\mathrm{cm}\).
\(\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{-3+2}{30}=-\dfrac{1}{30}\;\Rightarrow\;u=-30\,\mathrm{cm}\)
The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the ray diagram.

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer: Focal length of the convex mirror, \(f=+15\,\mathrm{cm}\); object distance, \(u=-10\,\mathrm{cm}\).
\(\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{15}+\dfrac{1}{10}=\dfrac{2+3}{30}=\dfrac{5}{30}=\dfrac16\;\Rightarrow\;v=+6\,\mathrm{cm}\)
The positive value of v indicates that the image is formed behind the mirror.
\(m=-\dfrac{v}{u}=-\dfrac{6}{-10}=+0.6\)
The positive value of magnification indicates that the image formed is virtual and erect (and diminished).

The magnification produced by a plane mirror is +1. What does this mean?

Answer: Magnification produced by a mirror is given by \(m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}\). The magnification produced by a plane mirror is +1. It shows that the image formed by the plane mirror is of the same size as that of the object. The positive sign shows that the image formed is virtual and erect.

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer: \(u=-20\,\mathrm{cm},\quad h=5\,\mathrm{cm},\quad R=30\,\mathrm{cm}\;\Rightarrow\;f=\dfrac{R}{2}=+15\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{15}+\dfrac{1}{20}=\dfrac{4+3}{60}=\dfrac{7}{60}\;\Rightarrow\;v=\dfrac{60}{7}=8.57\,\mathrm{cm}\)
The positive value of v indicates that the image is formed behind the mirror.
\(m=-\dfrac{v}{u}=-\dfrac{8.57}{-20}=+0.43\)
The positive value of magnification indicates that the image formed is virtual and erect.
\(h_2=m\,h_1=0.43\times5=2.14\,\mathrm{cm}\)
Therefore, the image formed is virtual, erect and smaller in size (2.14 cm).

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer: \(u=-27\,\mathrm{cm},\quad h=7\,\mathrm{cm},\quad f=-18\,\mathrm{cm}\)
\(\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{1}{18}+\dfrac{1}{27}=\dfrac{-3+2}{54}=-\dfrac{1}{54}\;\Rightarrow\;v=-54\,\mathrm{cm}\)
The screen should be placed at a distance of 54 cm in front of the given mirror.
\(m=-\dfrac{v}{u}=-\dfrac{-54}{-27}=-2\)
The negative value of magnification indicates that the image formed is real and inverted.
\(h_2=m\,h_1=-2\times7=-14\,\mathrm{cm}\)
The image is real, inverted and magnified, 14 cm in size.

Find the focal length of a lens of power −2.0 D. What type of lens is this?

Answer: \(P=\dfrac{1}{f\,(\text{in metres})}=-2\,\mathrm{D}\;\Rightarrow\;f=\dfrac{1}{-2}=-0.5\,\mathrm{m}\)
A concave lens has a negative focal length. Hence, it is a concave lens.

A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer: \(P=\dfrac{1}{f\,(\text{in metres})}=+1.5\,\mathrm{D}\;\Rightarrow\;f=\dfrac{1}{1.5}=\dfrac{10}{15}=+0.67\,\mathrm{m}\)
A convex lens has a positive focal length. Hence, it is a convex lens, or a converging lens.