CBSE Class 10 Science HOTs Reflection and Refraction Set 03

Question. Consider the situation where:
• An object is 3 cm (height).
• Mirror is concave with 6 cm focal length.
• Object is placed at the centre of curvature.
Which of the following options are correct? 
(a) The mirror will produce an image of magnification +1.5.
(b) The mirror will produce an image of magnification –1.
(c) The mirror will produce an image of magnification +1.
(d) The mirror will produce an image of magnification –1.5.
Answer: (b) The mirror will produce an image of magnification –1.

Question. An object is kept in front of a concave mirror of focal length 20 cm. The image is three times the size of the object. Calculate two possible distances of the object from the mirror.
Answer: Case 1: Real image (\( m = -3 \))
\( m = -\frac{v}{u} = -3 \implies v = 3u \)
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{-20} = \frac{1}{3u} + \frac{1}{u} = \frac{4}{3u} \)
\( 3u = -80 \implies u = -26.67 \text{ cm} \)

Case 2: Virtual image (\( m = +3 \))
\( m = -\frac{v}{u} = 3 \implies v = -3u \)
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{-20} = \frac{1}{-3u} + \frac{1}{u} = \frac{2}{3u} \)
\( 3u = -40 \implies u = -13.33 \text{ cm} \)

Question. An object is kept at a certain distance in front of a concave mirror with a focal length of 12 cm. The real image is formed at a point that is 10 cm away from the object. Calculate the magnification of the image formed by mirror.
Answer: Given \( f = -12 \text{ cm} \). For a real image, \( v \) and \( u \) are on the same side. Let \( u = -x \), then \( v = -(x + 10) \).
Using \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), we can solve for \( x \). Once \( u \) and \( v \) are found, magnification \( m = -\frac{v}{u} \).

Question. A student has focussed the image of an object of height 3 cm on a white screen using a concave mirror of focal length 12 cm. If the distance of the object from the mirror is 18 cm, find the values of following:
(i) Distance of the image from the mirror
(ii) Height of the image 
Answer: Given \( h_o = 3 \text{ cm} \), \( f = -12 \text{ cm} \), \( u = -18 \text{ cm} \).
(i) \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-12} - \frac{1}{-18} = \frac{-3 + 2}{36} = -\frac{1}{36} \implies v = -36 \text{ cm} \).
(ii) \( m = \frac{h_i}{h_o} = -\frac{v}{u} \implies h_i = -h_o \times \frac{v}{u} = -3 \times \frac{-36}{-18} = -6 \text{ cm} \).
The height of the image is 6 cm (inverted).

TOPIC : Uses of Spherical Mirrors

Question. In which of the following is a concave mirror used? 
(a) A solar cooker.
(b) A rear view mirror in vehicles.
(c) A safety mirror in shopping malls.
(d) In viewing full size image of distant tall buildings.
Answer: (a) A solar cooker.

Question. Which of the following mirror is used by a dentist to examine a small cavity?
(a) Convex mirror
(b) Plane mirror
(c) Concave mirror
(d) Combination of convex and concave mirror
Answer: (c) Concave mirror

Question. Magnification produced by the rear-view mirror fitted in vehicles
(a) is equal to one
(b) is greater than one
(c) is less than one
(d) depends on distance and height of the vehicle behind the driver’s vehicles.
Answer: (c) is less than one

Question. A full length image of the Taj Mahal can definitely be seen by using
(a) a plane mirror
(b) a concave mirror
(c) a convex mirror
(d) all of the options
Answer: (c) a convex mirror

Very Short Answer Type Questions 

Question. What should be the position of the object, when a concave mirror is used: 
(a) as a shaving mirror?
(b) in torches as reflecting mirror?
Answer: (a) Object should be placed between pole and focus.
(b) Object should be placed at focus.

Question. “Vehicles in this mirror are closer than they appear”. Why is this warning printed on the side view mirror of most vehicles? 
Answer: The phrase “vehicles in this mirror are closer than they appear” is a safety warning printed on the side view mirror of most vehicles. It is present because the side view mirror is a convex mirror and it gives a virtual image of wide field behind the vehicle and appears the objects smaller and farther away than they actually are, and the angular size of the virtual image is also smaller than the angular size of the object. During the lane change, a driver assumes that an adjacent vehicle is at a safe distance behind, when in fact it is quite a bit closer. The warning serves as a reminder to the driver of this potential problem.

Short Answer Type Questions 

Question. Name the type of mirror used (a) by dentist and (b) in vehicles headlights. Give reason why such mirrors are used in each case.
Answer: (a) Dentists use a concave mirror because it converges the light and when the object lies between its pole and principal focus, it forms a virtual, erect and enlarged image behind it. Hence they use the concave mirror so that they could see the cavity or plaque clearly, which is inside the teeth.
(b) Concave mirror is used in the vehicle headlights. The light bulb is placed at its focus. The light rays emitted from the bulb, after reflection from the mirror becomes parallel to the principal axis. This powerful parallel beam of light enabling the driver to see the road ahead clearly.

PRACTICE QUESTIONS

Question. In torches, search light and vehicles head lights the source of light is placed
(a) between the focus and centre of curvature of the reflector.
(b) at the centre of curvature of the reflector.
(c) very near to the focus of reflector.
(d) between pole and focus of the reflector.
Answer: (c) very near to the focus of reflector.

Question. If you look into the mirror and find that the image of your face is larger than your face, then the type of mirror is
(a) concave mirror
(b) convex mirror
(c) plane mirror
(d) combination of plane and concave mirror
Answer: (a) concave mirror

Question. Name the type of the mirrors used in design of solar furnaces. Explain how high temperature is achieved by this device.
Answer: Concave mirrors are used. They are large parabolic mirrors that concentrate sunlight (parallel rays from infinity) at their principal focus. The intense concentration of solar energy at the focus produces very high temperatures.

TOPIC : Refraction of Light

Question. The phenomenon of light exhibit in the case of apparent displacement of a pencil, partly immersed in water is
(a) refraction
(b) reflection
(c) dispersion
(d) scattering
Answer: (a) refraction

Question. As light travels from optically rarer medium to an optically denser medium, it will have
(a) increase in velocity
(b) decrease in velocity
(c) no change in velocity
(d) cannot say
Answer: (b) decrease in velocity

Question. A real image is formed by the light rays after reflection or refraction when they:
(A) actually meet or intersect with each other.
(B) actually converge at a point.
(C) appear to meet when they are produced in the backward direction.
(D) appear to diverge from a point.
Which of the above statements are correct? 
(a) (A) and (D)
(b) (B) and (D)
(c) (A) and (B)
(d) (B) and (C)
Answer: (c) (A) and (B)

Very Short Answer Type Questions

Question. For the following two cases, state whether the obliquely incident ray light on interface of medium 1 and medium 2 will bend towards or away from the normal after refraction, in medium 2.
Case I: (medium 1 having higher refractive index)
Case II: (medium 2 having higher refractive index)
Answer: Medium with higher refractive index will be optically denser medium with respect to the other. So obliquely incident light ray will:
Case I: bend away from the normal in medium 2 after refraction.
Case II: bend towards the normal in medium 2 after refraction.

PRACTICE QUESTIONS

Question. A lemon is kept in water in a glass tumbler, when viewed from the side appears to be
(a) bigger than its actual size
(b) smaller than its actual size
(c) same size as the as the actual size
(d) dark
Answer: (a) bigger than its actual size

Question. Choose the incorrect statement from the following:
(a) Letters appear to be raised, when seen through a glass slab placed over it.
(b) Light does not travel in the same direction in all media.
(c) Speed of light in different media is same.
(d) All of the options
Answer: (c) Speed of light in different media is same.

Question. For the same angle of incidence 45°, the angle of refraction in two transparent medium I and II are 20° and 30° respectively. Out of I and II, which medium is optically denser and why?
Answer: Medium I is optically denser. This is because for the same angle of incidence, the ray bends more towards the normal in medium I (angle of refraction 20°) than in medium II (angle of refraction 30°).

Refraction Through a Rectangular Glass Slab

Question. A light ray incident on a rectangular glass slab, after emerging from the opposite parallel faces, it is found that the
(a) \( \angle i > \angle e \)
(b) \( \angle i < \angle e \)
(c) \( \angle i = \angle e \)
(d) \( \angle i \neq \angle e \)A student has traced the path of
where \( \angle i = \text{angle of incidence} \)
\( \angle e = \text{angle of emergence} \)
Answer: (c) \( \angle i = \angle e \)

Question. Choose the correct statement for a rectangular glass slab.
(a) The emergent ray is not shifted sidewards slightly
(b) The emergent ray is perpendicular to the incident ray.
(c) The ray emerges parallel to the incident ray.
(d) The incident and emergent rays are inclined towards each other.
Answer: (c) The ray emerges parallel to the incident ray.

Question. A ray of light falling on a glass slab at 90°. The angle of incidence is
(a) 90°
(b) zero
(c) < 90° ≠ 0
(d) greater then 90°
Answer: (b) zero
When light rays incident along the normal, then by definition the angle of incidence is zero.

Question. Mohit took a thick glass slab whose one opposite parallel face is silvered. He observed that it formed multiple images. This happened on account of
(a) refraction of light
(b) reflection of light
(c) dispersion of light
(d) both (a) and (b)
Answer: (d) both (a) and (b)
Silvered face acts as a smooth reflecting surface.

Question. A student suggested the following ‘guidelines’ to his friend for doing the experiment on tracing the path of a ray of light passing through a rectangular glass slab for three different angles of incidence: [Delhi 2008]
A. Draw the ‘outline’ of the glass slab at three positions on the drawing sheet.
B. Draw ‘normals’ on the top side of these ‘outlines’ near their left end.
C. Draw the incident rays on the three ‘outlines’ in directions making angles of 30°, 45°, 60° with the normals drawn.
D. Fix two pins vertically on each of these incident rays at two points nearly 1 cm apart.
E. Look for the images of the ‘heads’ of these pins while fixing two pins from the other side, to get the refracted rays.
When he showed these ‘guidelines’ to his teacher, the teacher corrected and modified the ‘guidelines’ labelled as
(a) B, C, E
(b) B, D, E
(c) B, C, D
(d) C, D, E
Answer: (b) B, D, E
The normals are not generally at the ends. While fixing pins, the foot has to be taken as reference. The pins should be far off from each other.

Question. Select the correct statement(s) from the following regarding refraction of light when light is incident from the optically denser medium A on a optically rarer medium B.
I. Light will bend towards the normal in medium B.
II. Speed of light will be more in medium B as compared to medium A.
III. Angle of refraction will be more than angle of incidence.
IV. Angle of refraction will be less than angle of incidence.
(a) Both I and II
(b) Both II and III
(c) Both II and IV
(d) Only I, II and IV
Answer: (b) Both II and III

Very Short Answer Type Question

Question. (a) Why does the emergent ray in a glass slab is parallel to the incident ray?
(b) Why does the emergent ray shift sidewards?
Answer: (a) We know that, in case of refraction, when incident ray travels from air to glass, it bends towards normal and bends away from the normal when it travels through glass to air. Thus, the extent of bending of the ray of light at the opposite parallel faces i.e. air-glass and glass-air interface of rectangular glass slab are equal and opposite. Hence, the emergent ray is parallel to the incident ray in case of refraction through rectangular glass slab.
(b) The emergent ray shift sidewards slightly because of
(i) angle of incidence, and
(ii) thickness of glass slab

Short Answer Type Questions

Question. When a ray of light passes through the glass slab, then how many times does it change its path and why?
Answer: • When a ray of light passes through the glass slab, it bends twice.
• First time, at air-glass interface, it bends towards normal at denser medium i.e., inside glass slab. It is due to decrease in the speed of refracted light ray.
• Second time, at glass-air interface, the speed of light of refracted ray in rarer medium increases and the ray bends away from the normal.

Refractive Index

Question. Snell’s law can be derived from which type of incidence of light ray?
(a) Angle of refraction
(b) Angle of incidence
(c) Oblique incidence
(d) Angle of reflection
Answer: (c) Oblique incidence

Question. Which of the following represents correctly Snell’s law of refraction?
(a) \( \frac{\sin r}{\sin i} = \frac{1}{n} \)
(b) \( \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \)
(c) \( n_2 \sin i = \text{constant} \)
(d) All of these
Answer: (b) \( \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \)

Question. When an incident ray of light enters a medium from air, it bends towards the normal. Which of the following is TRUE about the refractive index of the medium (\( n_m \)) as compared to the refractive index of air (\( n_a \))? 
(a) \( n_m \) is equal to \( n_a \)
(b) \( n_m \) is less than \( n_a \)
(c) \( n_m \) is more than \( n_a \)
(d) The refractive indices cannot be compared based on the given information.
Answer: (c) The bending of the ray of light towards the normal indicates that the second medium is an optically denser medium whose refractive index is higher than that of air.

Question. The refractive index of medium A is 1.5 and that of medium B is 1.33. If the speed of light in air is \( 3 \times 10^8 \text{ m/s} \), what is the speed of light in medium A and B respectively? 
(a) \( 2 \times 10^8 \text{ m/s} \) and \( 1.33 \times 10^8 \text{ m/s} \)
(b) \( 1.33 \times 10^8 \text{ m/s} \) and \( 2 \times 10^8 \text{ m/s} \)
(c) \( 2.25 \times 10^8 \text{ m/s} \) and \( 2 \times 10^8 \text{ m/s} \)
(d) \( 2 \times 10^8 \text{ m/s} \) and \( 2.25 \times 10^8 \text{ m/s} \)
Answer: (d) Use \( v = \frac{c}{n} \).

Question. You are given three media A, B and C of refractive index 1.33, 1.65 and 1.46. The medium in which the light will travel fastest is
(a) A
(b) B
(c) C
(d) equal in all three media
Answer: (a) As, \( {}_a n_m = \frac{v_a}{v_m} \) or \( v_m = \frac{v_a}{{}_a n_m} \). So, the light will travel faster in a medium having lower refractive index.

Very Short Answer Type Questions

Question. A ray of light enters into benzene from air. If the refractive index of benzene is 1.50, by what percent does the speed of light reduce on entering the benzene?
Answer: The absolute refractive index of benzene is given by
\( n = \frac{\text{Speed of light in vacuum (c)}}{\text{Speed of light in benzene (v)}} \)
\( 1.50 = \frac{3 \times 10^8}{v} \)

\( \implies \) \( v = 2 \times 10^8 \text{ ms}^{-1} \)
Therefore, percentage decrease in the speed of light on entering the benzene.
\( = \frac{c - v}{c} \times 100\% \)
\( = \frac{3 \times 10^8 - 2 \times 10^8}{3 \times 10^8} \times 100 \)
\( = \frac{100}{3} = 33.33\% \)

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: Given: \( {}_{\text{air}} n_{\text{ice}} = 1.31 \), \( {}_{\text{air}} n_{\text{rock}} = 1.54 \)
\( {}_{\text{ice}} n_{\text{rock}} = \frac{{}_{\text{air}} n_{\text{rock}}}{{}_{\text{air}} n_{\text{ice}}} = \frac{1.54}{1.31} = 1.18 \)

Question. The refractive index of a medium ‘x’ with respect to ‘y’ is 2/3 and the refractive index of medium ‘y’ with respect to ‘z’ is 4/3. Calculate the refractive index of medium ‘z’ with respect of ‘x’. If the speed of light medium ‘x’ is \( 3 \times 10^8 \text{ ms}^{-1} \), calculate the speed of light in medium ‘y’. 
Answer: Given: \( n_{xy} = \frac{2}{3} \), \( n_{yz} = \frac{4}{3} \), \( n_{zx} = ? \)
We know that,
\( n_{xy} \times n_{yz} \times n_{zx} = 1 \)

\( \implies \) \( \frac{2}{3} \times \frac{4}{3} \times n_{zx} = 1 \)

\( \implies \) \( n_{zx} = \frac{9}{8} \)
Speed of light in medium ‘y’: Let the speed of light in medium y be \( v \) and speed of light in medium ‘x’ is \( c \), then \( n_{yx} = \frac{v}{c} \)

\( \implies \) \( \frac{1}{n_{xy}} = \frac{v}{c} \)

\( \implies \) \( \frac{1}{2/3} = \frac{v}{3 \times 10^8} \)

\( \implies \) \( v = 3 \times 10^8 \times \frac{2}{3} = 2 \times 10^8 \text{ ms}^{-1} \)
Hence, the speed of light in medium ‘y’ is \( 2 \times 10^8 \text{ ms}^{-1} \).

Question. Light enters from air to water having refractive index 4/3. Find the speed of light in water. The speed of light in vacuum is \( 3 \times 10^8 \text{ ms}^{-1} \).
Answer: Refractive index of water = \( \frac{4}{3} \)
Speed of light in vacuum = \( 3 \times 10^8 \text{ ms}^{-1} \)
Using, \( n = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in water}} = \frac{c}{v} \)

\( \implies \) \( \frac{4}{3} = \frac{3 \times 10^8}{v} \)

\( \implies \) \( v = \frac{9 \times 10^8}{4} = 2.25 \times 10^8 \text{ ms}^{-1} \)
∴ Speed of light in water = \( 2.25 \times 10^8 \text{ ms}^{-1} \).