CBSE Class 10 Science Light Reflection and Refraction VBQs Set 04

QUESTIONS

Question. Define the principal focus of a concave mirror.
Answer: The point on the principal axis, where all the incident rays, parallel to the principal axis, actually meet or converge after reflection from the concave mirror is called principal focus of a concave mirror.

Question. The radius of curvature of a spherical mirror is 20 cm. What is its focal length? 
Answer: We know \( f = \frac{R}{2} = \frac{20 \text{ cm}}{2} = 10 \text{ cm} \)
where \( f \) and \( R \) are the focal length and radius of curvature respectively.

Question. Name a mirror that can give an erect and enlarged image of an object.
Answer: Concave mirror.

Question. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer: Convex mirror is used because:
(i) it always produces a virtual and erect image.
(ii) the size of image formed is smaller than the object.
Therefore, it enables the driver to see a wide field view of the traffic behind the vehicle in a small mirror.

Question. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer: We know \( f = \frac{R}{2} = \frac{32}{2} = 16 \text{ cm} \)
Hence, the focal length of convex mirror is 16 cm.

Question. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer: Given: \( u = -10 \text{ cm} \), \( v = ? \), \( m = -3 \) (If image is real \( m \) is -ve).
Using the formula, \( m = -\frac{v}{u} \) we get,
\( \implies -3 = -\frac{v}{(-10)} \)
\( \implies v = -30 \text{ cm} \)
The image is formed at a distance of 30 cm from the mirror on the same side of the object.

Question. 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: A ray of light travelling in air enters obliquely into water, it will bend towards the normal. This is because the speed of light is lesser in a denser medium than the rarer medium. Water is optically denser than air.

Question. 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 \( 3 \times 10^8 \text{ ms}^{-1} \). 
Answer: The refractive index of the medium,
\( n_m = \frac{\text{Speed of light in air}}{\text{Speed of light in medium (glass)}} = \frac{c}{v_g} \)
Therefore, speed of light in glass,
\( v_g = \frac{c}{n_m} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ ms}^{-1} \).

Question. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer: This means that the ratio of speed of light in air to the speed of light in diamond is equal to 2.42.

Question. Define 1 dioptre of power of a lens.
Answer: 1 dioptre is the power of a lens whose focal length is 1 metre. \( 1\text{D} = 1 \text{ m}^{-1} \).

Question. 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: The size of real and inverted image is equal to the size of the object. This is possible only when the object is placed at \( 2F_1 \) of the convex lens. In this case, the image will also be formed at \( 2F_2 \) on the other side of the lens. This means, \( |u| = |v| = +50 \text{ cm} \) and \( v = 2f \).
Therefore, focal length of the convex lens is
\( f = \frac{v}{2} = \frac{50}{2} = +25 \text{ cm} = +0.25 \text{ m} \) [as \( v = 2f = +50 \text{ cm} \)]
Power of lens, \( P = \frac{1}{f \text{ (in metres)}} = \frac{1}{+0.25} = +4\text{D} \)
Hence, the power of the given lens is \( +4\text{D} \).

Question. Find the power of a concave lens of focal length 2 m.
Answer: Focal length of concave lens, \( f = -2 \text{ m} = -200 \text{ cm} \).
Here, negative sign is due to diverging nature of the concave lens.
Power of the lens, \( P = \frac{1}{f \text{ (in metres)}} = \frac{1}{-2} = -0.5 \text{D} \)
Hence, the power of the given lens is \( -0.5 \text{D} \).

NCERT EXERCISES

Question. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer: (d) Clay

Question. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer: (d) Between the pole of the mirror and its principal focus.

Question. Where should an object is placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer: (b) When an object is placed at \( 2F_1 \) of the convex lens, the image is formed at \( 2F_2 \) on the other side of the lens. The image formed is real, inverted and of the same size as the object.

Question. 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—
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer: (a) By sign convention, the negative sign of focal length indicates that it is the focal length of a concave mirror and a concave lens. Hence, both the spherical mirror and spherical lens are concave in nature.

Question. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) only either plane or convex.
Answer: (d) No matter whatever is the position of object, a convex mirror always forms a virtual, erect and diminished image of the object placed in front of it, whereas a plane mirror always forms a virtual, erect and of the same size image as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.

Question. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer: (c) For reading small letters, we need a lens which forms a virtual, erect and magnified image. This condition is fulfilled by a convex lens. Also, the magnification is more for a short focal length of lens. Hence, for reading small letters, a convex lens of focal length 5 cm would prefer to use.

Question. 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: Type of mirror used in
(a) Headlights of a car: Concave mirror. Concave mirror is used because light from the bulb placed at its focus gets reflected and a powerful parallel beam of light is obtained, which illuminates the road. This beam of light helps us to see things clearly upto a considerable distance at night.
(b) Side/rear-view mirror of a vehicle: Convex mirror. Convex mirror is used because (i) it always produces a virtual and erect image between its pole and focus. (ii) the size of image formed is smaller than the object irrespective of position of object. Therefore, it enables the driver to see wide field view of the traffic approaching from behind the vehicle in a small mirror.
(c) Solar furnace: Concave mirror. Concave mirror has the property to converge the sunlight along with heat radiation at its focus. As a result, the temperature at its focus increases and the substance placed at the focal point gets heated to a high temperature.

Question. 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 convex mirror, \( f = +15 \text{ cm} \)
Object distance, \( u = –10 \text{ cm} \)
According to the mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \text{, we get} \)
\( \frac{1}{15} = \frac{1}{v} + \frac{1}{-10} \)
\( \implies \) \( \frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{5}{30} = \frac{1}{6} \)
or \( v = +6 \text{ cm} \)
The positive value of \( v \) shows that the image is formed on the other side of the mirror, i.e. behind the mirror.
Magnification = \( -\frac{\text{Image distance}}{\text{Object distance}} \)
\( = -\frac{v}{u} = -\frac{6}{-10} = + 0.6 \)
The positive value of magnification shows that the image is virtual and erect.

Question. The magnification produced by a plane mirror is +1. What does it mean?
Answer: The positive sign in magnification shows that the image formed by a plane mirror is virtual, erect and exactly of the same size as that of the object.

Question. 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: For convex mirror,
\( h_o = + 5.0 \text{ cm}, u = –20 \text{ cm}, \)
\( f = \frac{R}{2} = \frac{30}{2} \text{ cm} = +15 \text{ cm} \)
Using mirror formula, \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \text{, we get} \)
\( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \)
\( \frac{1}{v} = \frac{1}{15} - \frac{1}{-20} = \frac{4 + 3}{60} \)
\( \implies \) \( v = \frac{60}{7} = + 8.57 \text{ cm} \)
Magnification, \( m = -\frac{v}{u} = \frac{h_i}{h_o} \text{ we get,} \)
\( h_i = 5 \times -\frac{8.57}{-20} = + 2.14 \text{ cm} \)
Since \( v \) is + ve, the image is virtual.
Since \( h_i = 2.14 \text{ cm} < 5.0 \text{ cm} \), i.e. \( h_i < h_o \), the image is diminished.
Hence a virtual, erect and diminished image of size 2.14 cm is formed behind the mirror at a distance of 8.57 cm from the pole of the convex mirror.

Question. 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: For concave mirror, \( h_o = 7.0 \text{ cm}, u = –27 \text{ cm}, f = –18 \text{ cm} \)
Using mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \text{, we get} \)
\( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{(-27)} \)
\( = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54} \)
\( \implies \) \( v = – 54 \text{ cm} \)
Hence the screen should be placed at a distance of 54 cm in front of the concave mirror to get the sharp focussed image of the object on it.
Magnification of spherical mirror is given by
\( m = \frac{h_i}{h_o} = -\frac{v}{u} \)
\( \implies \) \( h_i = -h_o \frac{v}{u} = -7 \times \frac{-54}{-27} = -14 \text{ cm} \)
The height of image is 14 cm.
Since \( h_i > h_o \), the image is enlarged.
Since \( v \) is –ve, the image is real and inverted.

Question. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer: Power of lens is given by
\( P = \frac{1}{f \text{ (in metres)}} \)
\( \implies \) \( -2.0 = \frac{1}{f} \)
\( \implies \) \( f = – 0.5 \text{ m} = – 50 \text{ cm} \)
Since the negative focal length of spherical lens indicates that it is concave in nature, hence, it is a concave lens.

Question. 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: Power of lens, \( P = \frac{1}{f \text{ (in metres)}} \)
\( \implies \) \( 1.5 = \frac{1}{f} \)
\( \implies \) \( f = \frac{1}{1.5} = \frac{2}{3} = + 0.67 \text{ m} = + 67 \text{ cm} \)
Hence the focal length of the given corrective lens is 0.67 m. The positive value of focal length indicates that the prescribed lens is converging, i.e. convex lens.

SELECT NCERT EXEMPLAR PROBLEMS

Question. A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
(a) – 30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm
Answer: (b) – 20 cm
Given \( h_o = +10 \text{ mm} = + 0.1 \text{ cm}, h_i = – 5 \text{ mm} = – 0.5 \text{ cm} \) for real image, \( v = – 30 \text{ cm} \)
Now, magnification, \( m = \frac{h_i}{h_o} = -\frac{v}{u} \)
On solving, we get, \( u = – 60 \text{ cm} \)
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = -\frac{1}{30} - \frac{1}{60} = -\frac{3}{60} = -\frac{1}{20} \)
\( \implies \) \( f = –20 \text{ cm} \)

Question. Which of the following statements is/are true?
(a) A convex lens has 4 dioptre power having a focal length 0.25 m
(b) A convex lens has –4 dioptre power having a focal length 0.25 m
(c) A concave lens has 4 dioptre power having a focal length 0.25 m
(d) A concave lens has – 4 dioptre having a focal length 0.25 m
Answer: (a) A convex lens has 4 dioptre power having a focal length 0.25 m
Positive sign with power and focal length indicates that the given lens is convex.
Also \( f = \frac{1}{P} = \frac{1}{4} = 0.25 \text{ m} \)

Question. Magnification produced by a rear view mirror fitted in vehicles
(a) is less than one
(b) is more than one
(c) is equal to one
(d) can be more than or less than one depending upon the position of the object in front of it.
Answer: (a) is less than one
Convex mirror is used as rear view mirror and always forms virtual, erect and diminished image. So magnification produced by a rear view mirror is less than one.

Question. The laws of reflection hold good for
(a) plane mirror only
(b) concave mirror only
(c) convex mirror only
(d) all mirrors irrespective of their shape
Answer: (d) all mirrors irrespective of their shape
The laws of reflection hold good for light reflected from any smooth surface irrespective of their shapes.

Question. In which of the following, the image of an object placed at infinity will be highly diminished and point sized?
(a) Concave mirror only
(b) Convex mirror only
(c) Convex lens only
(d) Concave mirror, convex mirror, concave lens and convex lens.
Answer: (d) Concave mirror, convex mirror, concave lens and convex lens.
The incident ray coming from the object placed at infinity will be parallel to the principal axis. When the parallel beam of light incident on a mirror or lens, irrespective of their nature, after reflection/refraction, will pass or appear to pass through their principal focus. Hence highly diminished and point size image will be formed at their focus.

Question. Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.
(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.
(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.
Answer: (a) Concave mirror
(b) Convex lens
(c) Concave lens
(d) Convex mirror

Question. A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appears to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason. 
Answer: No, it will be dependent on the refractive index of the liquid in which the pencil is dipped. The tip at the bottom will get elevated as per the relation.
Refractive index (\( n \)) = \( \frac{\text{real depth}}{\text{apparent depth}} \)

Question. Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.
Answer: Given: \( n_{dg} = 1.6, n_g = 1.5 \)
Since \( n_{dg} = \frac{n_d}{n_g} \)
We have \( n_d = n_g \times n_{dg} = 1.5 \times 1.6 = 2.4 \)

Question. A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Answer: Yes, for getting a virtual image, the object has to be placed between the optical centre and the focus. For a real and enlarged image, the object should be between \( F_1 \) and \( 2F_1 \).

Question. Sudha finds out that a sharp image of the windowpane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the windowpane without disturbing the lens. In which direction will she move the screen to obtain the sharp image of the building? What is the approximate focal length of this lens? 
Answer: She will move the screen towards the lens such that the final image is at the focus. The focal length will be slightly less than 15 cm as building can be treated as the object at infinite distance.

Question. How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?
Answer: Power of a lens is defined as the reciprocal of the focal length.
\( P = \frac{1}{f \text{ (in m)}} = \frac{100}{f \text{ (in cm)}} \)
\( \implies \) \( P_{20} = \frac{100}{20} = 5\text{D} \)
\( P_{40} = \frac{100}{40} = 2.5\text{D} \)
So, the lens of 20 cm focal length will converge more.

Question. The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
Answer: Since \( m = \frac{h_i}{h_o} = \frac{v}{u} \) for a lens.
Since Image is real.
\( \implies \) \( m = -3 = \frac{v}{u} \)
With \( v = 80 \text{ cm}, u = \frac{v}{-3} = -\frac{80}{3} \text{ cm} \). The image is obtained on the screen placed on the other side of the lens, it is possible only in the case of convex lens.
Hence, the candle should be placed at a distance of \( \frac{80}{3} = 26.6 \text{ cm} \) to the left of the convex lens to form the real and inverted image at a distance of 80 cm on the right side of the lens.

Question. The size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
Answer: If the image is real (as in concave mirror),
\( m = -\frac{1}{3} \). Therefore, \( -\frac{v}{u} = -\frac{1}{3} \).
Using mirror formula, \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), we get
\( \frac{1}{-20} = \frac{1}{u/3} + \frac{1}{u} = \frac{3}{u} + \frac{1}{u} = \frac{4}{u} \)
\( \implies \) \( u = -80 \text{ cm} \)

If the image is virtual (as in convex mirror),
\( m = \frac{1}{3} \). Therefore, \( -\frac{v}{u} = \frac{1}{3} \).
Using mirror formula, \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), we get
\( \frac{1}{20} = \frac{1}{-u/3} + \frac{1}{u} = \frac{1}{u}(1 - 3) = \frac{-2}{u} \)
\( \implies \) \( u = -40 \text{ cm} \)