CBSE Class 10 Science Light Reflection and Refraction VBQs Set 02

Short Answer Type Questions

Question. (a) For the same angle of incidence in media P, Q and R, the angles of refraction are 45°, 35° and 15° respectively. In which medium will the velocity of light be (i) minimum (ii) maximum? Give reason for your answer. 
(b) When light enters from air to glass, the angles of incidence and refraction in air and glass are 45° and 30° respectively. Find the refractive index of glass. (Given that \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), \( \sin 30^\circ = \frac{1}{2} \)) 
Answer: (a) From Snell’s law, \( n = \frac{\sin i}{\sin r} \)
Also, \( n = \frac{c}{v} \), where \( c = \text{velocity of light in vacuum /air} \)
\( \therefore \frac{c}{v} = \frac{\sin i}{\sin r} \)
Since \( c \) and \( \sin i \) are constant. Therefore, \( \sin r \propto v \).
Since the value of \( \sin r \) will be least for 15° and maximum for 45° among the three, the velocity of light will be (i) minimum in medium R. (ii) maximum in P.
(b) Using Snell’s law,
\( n_{ga} = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ} \)

\( \implies \) \( n_{ga} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} = 1.41 \)

Question. State the laws of refraction of light. Explain the term ‘absolute refractive index of a medium’ and write an expression to relate it with the speed of light in vacuum. 
Answer: Laws of refraction of light:
(a) The incident ray, the normal at the point of incidence and the refracted ray, all lie in the same plane for the two given transparent media.
(b) The ratio of sine of angle of incidence, i.e. \( \sin i \) to the sine of angle of refraction, i.e. \( \sin r \) is always constant, for the light of a given colour and for the given pair of media.
Mathematically, \( \frac{\sin i}{\sin r} = \text{constant} = n_{21} \)
The constant \( n_{21} \) is called the refractive index of the second medium with respect to the first medium.
Absolute Refractive Index: The refractive index of medium 2 with respect to vacuum or air is considered to be its absolute refractive index. It is represented by \( n_2 \). It is also equal to the speed of light in vacuum to the speed of light in the medium.
i.e. \( n_2 = \frac{\text{Speed of light in air or vacuum (c)}}{\text{Speed of light in the medium (v)}} \)
\( = \frac{c}{v} \)

Question. (a) Find the absolute refractive index of a medium in which light travels with a speed of \( 1.4 \times 10^8 \text{ m/s} \).
(b) How do we distinguish a medium to be rarer or denser? Give two reasons.
Answer: (a) Absolute refractive index of the medium is given by
\( n_m = \frac{\text{Speed of light in vacuum (c)}}{\text{Speed of light in medium (v)}} \)
i.e. \( n_m = \frac{c}{v} \)
Given: \( c = 3 \times 10^8 \text{ m/s} \), \( v = 1.4 \times 10^8 \text{ m/s} \)
\( \therefore n_m = \frac{c}{v} = \frac{3 \times 10^8}{1.4 \times 10^8} = \frac{3}{1.4} = 2.14 \)
(b) (i) Based on the bending of light.
(ii) Based on the velocity of light in the medium or by knowing its refractive index.

Long Answer Type Question

Question. (a) On entering in a medium from air, the speed of light becomes half of its value in air. Find the refractive index of that medium with respect to air?
(b) A glass slab made of a material of refractive index \( n_1 \) is kept in a medium of refractive index \( n_2 \). A light ray is incident on the slab. Draw the path of the rays of light emerging from the glass slab, if (i) \( n_1 > n_2 \) (ii) \( n_1 = n_2 \) (iii) \( n_1 < n_2 \). 
Answer: (a) Refractive Index of a medium (\( n \))
\( = \frac{\text{Velocity of light in vacuum}}{\text{Velocity of light in the medium}} \)
Let the velocity of light in vacuum be \( v_1 \) and velocity of light in the medium be \( v_2 \).
\( \frac{v_1}{2} = v_2 \)
Hence \( n = \frac{v_1}{v_2} \)

\( \implies \) \( n = \frac{v_1}{(v_1/2)} = 2 \)
(b) (i) The ray moves towards the normal. [Diagram shows ray bending towards normal inside \( n_1 \)].
(ii) The ray moves undeviated. [Diagram shows straight path through \( n_1 \) and \( n_2 \)].
(iii) The ray moves away from the normal. [Diagram shows ray bending away from normal inside \( n_1 \)].

PRACTICE QUESTIONS

Question. When light is incident on a glass slab, the incident ray, refracted ray and the emergent ray are in the three media A, B and C. If \( n_1 \), \( n_2 \) and \( n_3 \) are the refractive indices of A, B and C respectively and the emergent ray is parallel to the incident ray, which of the following is true?
(a) \( n_1 < n_2 < n_3 \)
(b) \( n_1 > n_2 > n_3 \)
(c) \( n_1 < n_2 = n_3 \)
(d) \( n_1 = n_3 < n_2 \)
Answer: (d) \( n_1 = n_3 < n_2 \)

Question. Prove that refraction will not take place at the boundary that separates two media of equal refractive indices. Draw ray diagram to justify this statement.
Answer: Refraction involves a change in the speed of light and bending of its path when it enters from one medium to another of different optical density. According to Snell's law, \( \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \). If \( n_1 = n_2 \), then \( \sin i = \sin r \), which means \( \angle i = \angle r \). Thus, the light ray continues in a straight line without bending. [Ray diagram showing a straight light path across the boundary].

Question. With respect to air, the refractive index of ice is 1.31 and that of rock salt is 1.54. Calculate the refractive index of rock salt with respect to ice.
Answer: \( {}_{\text{ice}} n_{\text{rock salt}} = \frac{{}_{\text{air}} n_{\text{rock salt}}}{{}_{\text{air}} n_{\text{ice}}} = \frac{1.54}{1.31} = 1.18 \).

Refraction by Spherical Lenses and Formation of Images

Question. The following diagram shows the use of an optical device to perform an experiment of light. As per the arrangement shown, the optical device is likely to be a; 
(a) Concave mirror
(b) Concave lens
(c) Convex mirror
(d) Convex lens
Answer: (b) Concave lens

Question. A divergent lens will produce
(a) always real image
(b) always virtual image
(c) both real and virtual image
(d) None of the options
Answer: (b) always virtual image

Question. When object moves closer to convex lens, the image formed by it shift
(a) away from the lens
(b) towards the lens
(c) first towards and then away from the lens
(d) first away and then towards the lens
Answer: (a) away from the lens

Question. When object moves closer to a concave lens the image formed by it shifts
(a) away from the lens on the same side of object
(b) toward the lens
(c) away from the lens on the other side of lens
(d) first towards and then away from the lens
Answer: (b) toward the lens

Question. State the two factors on which focal length of the spherical lens depends.
Answer: Focal length of spherical lens depends on the
(i) refractive index of the glass, and
(ii) radius of curvature of its two surfaces.

Question. The images formed by an ordinary convex lens suffer from a defect, called chromatic defect, which leads to false coloured edges in the images. This happens because light rays of different colours bend differently as they enter and leave the lens. If a parallel white light beam passes through a convex lens, the light of which colour (among violet to red in the spectrum) will converge at a point closest to the lens? Justify your answer.
Answer:

• The violet colour will converge at a point closest to the lens.
• Among the colour components of white light in the spectrum, glass has highest refractive index for violet colour light. Hence, the deviation of violet light will be maximum after refraction through the glass lens.

Short Answer Type Questions

Question. Draw a ray diagram to show the path of the refracted ray in each of the following cases: A ray of light incident on a concave lens is (a) passing through its optical centre. (b) parallel to its principal axis. (c) directed towards its principal focus. 
Answer: (a) [Ray diagram showing incident ray passing straight through the optical centre O of a concave lens].
(b) [Ray diagram showing incident ray parallel to principal axis, diverging after refraction such that it appears to come from the focus F on the same side].
(c) [Ray diagram showing incident ray directed towards the principal focus F on the opposite side, becoming parallel to the principal axis after refraction].

PRACTICE QUESTIONS

Question. Study the diagram given below and identify the type of lens XX′ and the position of the point on the principal axis OO′ where the image of the object AB appear to be formed: 
(a) Concave; between O′ and Y
(b) Concave; between O and Y
(c) Convex; between O′ and Y
(d) Convex; between O and Y
Answer: (c) Convex; between O′ and Y

Question. A student has focused on the screen a distant building using a convex lens. If he has selected a blue coloured building as object, select from the following options the one which gives the correct characteristics of the image formed on the screen. 
(a) virtual, erect, diminished and in green shade
(b) real, inverted, diminished and in violet shade
(c) real, inverted, diminished and in blue shade
(d) virtual, inverted, diminished and in blue shade
Answer: (c) real, inverted, diminished and in blue shade

Question. How does the focal length of a thick convex lens differ from that of a thin lens made of the same glass?
Answer: A thick convex lens has a shorter focal length compared to a thin lens because it has a greater curvature and therefore more converging power.

Question. What is the difference between a convex lens and a biconvex lens?
Answer: A convex lens is a general term for any lens that is thicker in the middle, while a biconvex lens specifically refers to a lens where both surfaces are curved outwards (spherical surfaces bulging outwards).

Question. Which lens bends more light, thick convex lens or thin convex lens made of the same material?
Answer: A thick convex lens bends light more because it has higher refractive power due to its greater thickness and curvature.

Question. (a) On what factor does the size of image formed by a thin convex lens depends? (b) A ray of light is passing through the principal focus of a convex lens. How will it emerge after refraction through the lens? (c) Which law is obeyed by the light rays when refracted through a spherical lens?
Answer: (a) The size of the image depends on the position of the object with respect to the focal length of the lens.
(b) It will emerge parallel to the principal axis.
(c) Snell's Law of refraction.

Question. (a) What is the focal length of the lens used in sunglasses? (b) The following figures show the path of light rays through three lenses marked \( L_1 \), \( L_2 \) and \( L_3 \) and their focal points \( F_1 \), \( F_2 \) and \( F_3 \) respectively. Identify the nature of lenses.
Answer: (a) Sunglasses are usually made of zero-power lenses, so their focal length is infinite.
(b) (i) Convex lens (converging), (ii) Concave lens (diverging), (iii) Convex lens (converging).

TOPIC : Sign Convention for Spherical Lenses, Lens Formula and Magnification

Multiple Choice Questions 

Question. For the diagram shown, according to New Cartesian Sign Convention, the sign of object distance, image distance and focal length for the given lens will be
(a) – , + , +
(b) – , – , +
(c) – , – , –
(d) + , – , +
Answer: (a) – , + , +

Question. If the magnification produced by lens has a negative value, the image will be
(a) real and erect
(b) virtual and erect
(c) real and inverted
(d) virtual and inverted
Answer: (c) real and inverted

Question. Linear magnification of a concave lens is always positive but less than one. This is because it forms
(a) real image only
(b) virtual and erect image on the other side of lens
(c) virtual and diminished image on the same side of object irrespective of its position.
(d) virtual, erect and enlarged image
Answer: (c) virtual and diminished image on the same side of object irrespective of its position.

Question. A converging lens forms a three times magnified image of an object, which can be taken on a screen. If the focal length of the lens is 30 cm, then the distance of the object from the lens is 
(a) –55 cm
(b) –50 cm
(c) –45 cm
(d) –40 cm
Answer: (d) For real image, \( m = -3 = \frac{v}{u} \)
\( \implies v = - 3u \). By lens formula, we get \( u = -40 \text{ cm} \).

Question. The condition for which the spherical lens has equal focal length on either side of the lens is 
(a) same medium on either side of lens
(b) same radius of curvature of both curved surface
(c) independent of medium on either side of lens.
(d) both (a) and (b)
Answer: (d) both (a) and (b)

Question. An object is placed at distance of 15 cm in front of a concave lens of focal length 15 cm. The position of image formed will be at a distance of
(a) – 15 cm
(b) + 15 cm
(c) – 7.5 cm
(d) + 7.5 cm
Answer: (c) Substitute \( u = -15 \text{ cm} \), \( f = -15 \text{ cm} \) in lens formula, we get, \( v = -7.5 \text{ cm} \).

Question. A teacher sets up the stand carrying a convex lens of focal length 21 cm placed at 42 cm mark on the optical bench. He asks four students A, B, C and D to suggest the position of screen on the optical bench so that a distinct image of a distance tree obtained on it. The correct position of the screen on the optical bench suggested by one of the student is
(a) 63 cm
(b) 32 cm
(c) 21 cm
(d) 84 cm
Answer: (a) The parallel rays coming from the distant tree are focused by the convex lens at its focal point. Therefore, the position of screen on optical bench is \( 42 + 21 = 63 \text{ cm} \).

Very Short Answer Type Questions

Question. A real image 2/3rd of the size of the object is formed by a convex lens when the object is at a distance of 12 cm from it. Find the focal length of the lens. 
Answer: Given: \( h' = -\frac{2}{3} h \), \( u = -12 \text{ cm} \)
• Magnification, \( m = \frac{h'}{h} = \frac{v}{u} \)
\( \implies v = \frac{h'}{h} \times u = \frac{-2/3 h}{h} \times (-12) = 8 \text{ cm} \)
• Using lens formula
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{8} - \frac{1}{-12} = \frac{1}{8} + \frac{1}{12} = \frac{5}{24} \)
\( \therefore f = \frac{24}{5} = +4.8 \text{ cm} \)

Question. State the meaning of linear magnification. How is it related to object distance and image distance? When is magnification positive or negative?
Answer: Linear magnification is the ratio of the height of the image to the height of the object. It is represented by the letter ‘m’.
\( m = \frac{\text{Height of image } (h_i)}{\text{Height of object } (h_o)} = \frac{v}{u} \)
where \( h_i \) is the height of the image and \( h_o \) is the height of the object. If the image formed is virtual and erect, then the magnification is positive and if the image formed is real and inverted, then the magnification is negative.

Short Answer Type Questions

Question. An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm. 
Answer: Given: \( h_o = 6 \text{ cm} \), \( f = -5 \text{ cm} \), \( u = -10 \text{ cm} \)
Using lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \therefore -\frac{1}{5} = \frac{1}{v} - \frac{1}{-10} \)
or \( \frac{1}{v} = -\frac{1}{5} - \frac{1}{10} = \frac{-2 - 1}{10} = \frac{-3}{10} \)
\( v = -\frac{10}{3} \text{ cm} \)
Thus, the image is formed on the same side of the object at a distance of \( \frac{10}{3} \text{ cm} \) from the optical centre of the lens. The negative sign indicates that the image is virtual.
Using the formula, \( m = \frac{h_i}{h_o} = \frac{v}{u} \)
\( \implies h_i = \frac{v}{u} \times h_o = \frac{-10/3}{-10} \times 6 = +2 \text{ cm} \)
The positive sign indicates that the image is erect.

Question. An object placed on a metre scale at 8 cm mark was focussed on a white screen placed at 92 cm mark, using a converging lens placed on the scale at 50 cm mark. (a) Find the focal length of converging lens. (b) Find the position of the image formed if the object is shifted towards the lens at a position of 29.0 cm. (c) State the nature of the image formed if the object is further shifted towards the lens. 
Answer: (a) \( u = -(50 - 8) = -42 \text{ cm} \), \( v = 92 - 50 = 42 \text{ cm} \)
Focal length of converging lens (convex lens) is given by \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) (lens formula)
\( \frac{1}{f} = \frac{1}{42} - \frac{1}{-42} = \frac{2}{42} = \frac{1}{21} \)
\( \therefore f = 21 \text{ cm} \)
(b) Now, the object is shifted towards the lens at a position of 29.0 cm. Therefore, new object distance, \( u' = -(50 - 29.0) = -21 \text{ cm} \).
Again lens formula, \( \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \)
\( \frac{1}{21} = \frac{1}{v'} - \frac{1}{-21} = \frac{1}{v'} + \frac{1}{21} \)
\( \therefore \frac{1}{v'} = \frac{1}{21} - \frac{1}{21} = 0 \)
or \( v' = \frac{1}{0} = \infty \)
So, the image will be formed at infinity.
(c) If the object is further shifted towards the lens, the object is now within the focus of a convex lens so the nature of image formed is

• virtual and erect, and
• enlarged or magnified, i.e. larger than the size of the object.

Question. An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image-distance and height of the image formed.
Answer: Given: For converging lens, object distance (\( u \)) = –25 cm; Focal length (\( f \)) = 15 cm; height of object (\( h_o \)) = 10 cm
Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Substituting \( u \) and \( f \), we get
\( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{15} + \frac{1}{-25} = \frac{2}{75} \)
\( v = 37.5 \text{ cm} \)
Hence, image distance = 37.5 cm
Magnification of the lens is given by \( m = \frac{h_i}{h_o} = \frac{v}{u} \)
\( h_i = h_o \times \frac{v}{u} = 10 \times \frac{37.5}{-25} = -15 \text{ cm} \)
The position of image is at distance 37.5 cm from the lens. The image is formed on the right side of the lens and it is real and inverted in nature.

Question. (a) A lens of focal length 5 cm is being used by Debashree in the laboratory as a magnifying glass. Her least distance of distinct vision is 25 cm. What is the magnification obtained by using the glass? (b) Ravi kept a book at a distance of 10 cm from the eyes of his friend Hari. Hari is not able to read anything written in the book. Give reasons for this? 
Answer: (a) Given, image distance = \( v = -25 \text{ cm} \), focal length = \( f = 5 \text{ cm} \), magnification = \( m = ? \)
From lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \)
\( \implies \frac{1}{u} = \frac{1}{-25} - \frac{1}{5} = \frac{-1 - 5}{25} = \frac{-6}{25} \)
Object distance = \( u = \frac{-25}{6} \text{ cm} \)
We know that, \( m = \frac{v}{u} = \frac{-25}{-25/6} = 6 \).
(b) This is because the least distance of distinct vision is 25 cm.

Long Answer Type Questions

Question. (a) Explain the following terms related to spherical lenses: (i) Optical centre (ii) Aperture (iii) Focal length (b) A converging lens has focal length of 12 cm. Calculate at what distance should the object be placed from the lens so that it forms an image at 48 cm on the other side of the lens.
Answer: (a) (i) Optical centre: The centre point of a lens is known as its optical centre. It always lies inside the lens. A light beam passing through the optical centre emerges out without any deviation after refraction.
(ii) Aperture: This is the length of the lens through which refraction takes place.
(iii) Focal length: The distance of the principal focus from the optical centre of the spherical lens is called the focal length (\( f \)) of the lens.
(b) Focal length of the converging lens, \( f = + 12 \text{ cm} \). Image distance, \( v = + 48 \text{ cm} \) (+ ve sign is taken because of sign convention). Using lens formula, \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \implies \frac{1}{12} = \frac{1}{48} - \frac{1}{u} \)
\( \therefore \frac{1}{u} = \frac{1}{48} - \frac{1}{12} = \frac{-3}{48} \)
\( u = -16 \text{ cm} \).
So, the distance of the object from the lens is 16 cm.

PRACTICE QUESTIONS

Question. If the real image of the candle flame formed by a lens 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? 
(a) –80 cm
(b) –40 cm
(c) –40/3 cm
(d) –80/3 cm
Answer: (d) –80/3 cm

Question. A concave lens has focal length of 20 cm. At what distance from the lens 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: Given \( f = -20 \text{ cm}, v = -15 \text{ cm} \). Using lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), we find \( u = -60 \text{ cm} \). Magnification \( m = \frac{v}{u} = \frac{-15}{-60} = \frac{1}{4} \). Image size \( h_i = m \times h_o = \frac{1}{4} \times 5 = 1.25 \text{ cm} \).