CBSE Class 10 Science HOTs Reflection and Refraction Set 07

Case Based Questions

Read the passage given below and answer the following questions :
The curved surface of a spoon can be considered as a spherical mirror. A highly smooth polished surface is called mirror. The mirror whose reflecting surface is curved inwards or outwards is called a spherical mirror. Inner part works as a concave mirror and the outer bulging part acts as a convex mirror. The center of the reflecting surface of a spherical mirror is called pole and the radius of the sphere of which the mirror is formed is called radius of curvature.

Question. When a concave mirror is held towards the sun and its sharp image is formed on a piece of carbon paper for some time, a hole is burnt in the carbon paper. What is the name given to the distance between the mirror and carbon paper?
Answer: The name given to the distance is the focal length of the concave mirror. This is because the Sun is at infinity, and its rays are parallel to the principal axis, which converge at the principal focus after reflection.

Question. On what factors does the focal length of a spherical mirror depend?
Answer: The focal length of a spherical mirror depends only on its radius of curvature \( (R) \). It is given by the relation \( f = R/2 \).

Question. Name the type of mirror used in the designing of solar furnaces. Explain how can high temperature is achieved by this device.
Answer: Concave mirrors are used in the design of solar furnaces. High temperature is achieved because concave mirrors are converging mirrors; they concentrate the parallel rays of the Sun falling on their large surface area to a single point called the principal focus. The intense concentration of solar energy at the focus produces a high temperature.

Question. List two possible ways in which a concave mirror can produce a magnified image of an object placed in front of it. State the difference if any between these two images.
Answer: A concave mirror can produce a magnified image in two ways:
(i) When the object is placed between the pole \( (P) \) and the principal focus \( (F) \). The image formed is virtual, erect, and magnified.
(ii) When the object is placed between the principal focus \( (F) \) and the centre of curvature \( (C) \). The image formed is real, inverted, and magnified.
The main difference is that the first case produces a virtual and erect image, while the second case produces a real and inverted image.

A & R Questions

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question. Assertion : If a ray of light is incident on a convex mirror along its principal axis, then the angle of incidence as well as the angle of reflection for a ray of light will be zero.
Reason : A ray of light going towards the centre of curvature of a convex mirror is reflected back along the same path.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Question. Assertion : Light is able to reach the earth from the sun.
Reason : Light rays can travel in vaccum.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Question. Assertion : A convex lens is made of two different materials. A point object is placed on the principal axis. The number of images formed by the lens will be two.
Reason : The image formed by convex lens is always virtual.
Answer: (c) Assertion is correct statement but reason is wrong statement.

Question. Assertion : In the case of concave mirror, the minimum distance between real object and its real image is zero.
Reason : If concave mirror forms virtual image of real object, the image is magnified.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Multiple Choice Questions

Question. Light travels from air into glass of refractive index 1.5. The time taken by the light to travel through a piece of glass of 50 cm thickness is
(a) 2.25 s
(b) \( 2.25 \times 10^{-7} \) s
(c) \( 2.5 \times 10^{-8} \) s
(d) \( 2.5 \times 10^{-9} \) s
Answer: (d) \( 2.5 \times 10^{-9} \) s

Question. To form an image twice the size of the object, using a convex lens of focal length 20 cm, the object distance must be
(a) < 20 cm
(b) > 20 cm
(c) < 20 cm and between 20 cm and 40 cm
(d) cannot say
Answer: (c) < 20 cm and between 20 cm and 40 cm

Question. An object is at a distance of 10 cm from a mirror and the image of the object is at a distance of 30 cm from the mirror on the same side as the object. Then the nature of the mirror and its focal length is
(a) convex, 15 cm
(b) concave, 1.5 cm
(c) convex, 7.5 cm
(d) concave, 7.5 cm.
Answer: (d) concave, 7.5 cm.

Question. With regard to refraction which of the following statement is false ?
(a) It is a change in direction of light when it passes from one transparent medium into another of different optical density.
(b) Light is deviated away from the normal when it enters an optically dense medium from a less dense medium.
(c) The velocity of light is changed during refraction.
(d) The wavelength of the light is changed during refraction.
Answer: (b) Light is deviated away from the normal when it enters an optically dense medium from a less dense medium.

Question. A man 180 cm high stands in front of a plane mirror. His eyes are at a height of 172 cm from the floor. Then to see his full image for minimum length of mirror, the lower end of the mirror should be placed at a height of
(a) 86 cm from the floor
(b) 94 cm from the floor
(c) 4 cm from the floor
(d) 8 cm from the floor
Answer: (a) 86 cm from the floor

Question. A convex lens is made of a material having refractive index 1.2. If it is dipped in water \( (\mu = 1.33) \), it will behave like a
(a) convergent lens
(b) divergent lens
(c) a rectangular slab
(d) a prism
Answer: (b) divergent lens

VSA Type Questions

Question. State the expression for lateral magnification of a concave mirror in terms of object distance and image distance.
Answer: Lateral magnification \( (m) \) of a concave mirror is given by the ratio of image distance \( (v) \) to the object distance \( (u) \) with a negative sign:
\( m = -\frac{v}{u} \)

Question. A tank of water is 4 m deep. How deep does it appear when seen normally ?
Answer: For normal incidence, apparent depth \( d' = \frac{\text{Real depth } d}{\text{Refractive index } \mu} \). Assuming \( \mu \text{ for water} \approx 1.33 \):
\( d' = \frac{4}{1.33} \approx 3 \text{ m} \).
The tank appears to be about 3 m deep.

Question. Define aperture of a mirror.
Answer: The diameter of the reflecting surface of a spherical mirror is called its aperture.

Question. According to the sign convention, which mirror has negative focal length?
Answer: According to the Cartesian sign convention, a concave mirror has a negative focal length.

SA I Type Questions

Question. Rohit while playing with an old lens discovers that, if he holds the lens 20 cm away from a wall opposite to a window, he can see sharp but inverted images of outside world on the wall. What is the power of the old lens holded by him?
Answer: Since the lens forms a sharp image of a distant object (outside world) on the wall, the wall must be at the principal focus of the lens. Therefore, the focal length \( (f) \) of the lens is 20 cm or 0.2 m.
Power \( (P) = \frac{1}{f(\text{in m})} = \frac{1}{0.2} = +5 \text{ D} \).
The power of the lens is +5 D.

Question. How can you distinguish between plane mirror, convex mirror and concave mirror by merely looking at the image formed in each case?
Answer: By looking at the image of our face when the mirror is held close:
1. If the image is of the same size and erect, it is a plane mirror.
2. If the image is virtual, erect, and magnified, it is a concave mirror.
3. If the image is virtual, erect, and diminished, it is a convex mirror.

Question. What change do you expect in the focal length of a concave mirror and a convex lens, when immersed in water?
Answer: The focal length of a concave mirror remains unchanged when immersed in water as it depends only on the radius of curvature. However, the focal length of a convex lens increases when immersed in water because its refractive index with respect to water is less than its refractive index with respect to air.

Question. Two media with refractive indices 1.31 and 1.50 are given. In which case, (i) Bending of light is more? (ii) Speed of light is more?
Answer: (i) Bending of light is more in the medium with the higher refractive index, which is 1.50.
(ii) Speed of light is more in the medium with the lower refractive index, which is 1.31 \( (v = c/n) \).

SA II Type Questions

Question. An object is 8 cm high. It is desired to form a real image 4 cm high at 60 cm from the mirror. What is focal length and the type of mirror needed?
Answer: Given \( h_o = 8 \text{ cm} \), \( h_i = -4 \text{ cm} \) (real image is inverted), and \( v = -60 \text{ cm} \) (real image).
Magnification \( m = \frac{h_i}{h_o} = \frac{-4}{8} = -0.5 \).
Also, \( m = -\frac{v}{u} \)
\( -0.5 = -\frac{-60}{u} \)

\( \implies \) \( -0.5 = \frac{60}{u} \)

\( \implies \) \( u = \frac{60}{-0.5} = -120 \text{ cm} \).
Using mirror formula: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
\( \frac{1}{f} = \frac{1}{-60} + \frac{1}{-120} = \frac{-2 - 1}{120} = -\frac{3}{120} = -\frac{1}{40} \)

\( \implies \) \( f = -40 \text{ cm} \).
Since the focal length is negative and it forms a real image, the mirror needed is a concave mirror.

LA Type Questions

Question. In going from a rarer to a denser medium, a ray of light bends towards normal. And in going from a denser to a rarer medium, a ray of light bends away from normal. Explain why.
Answer: This is explained by Snell's law: \( n_1 \sin i = n_2 \sin r \) or \( \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \).
1. From rarer \( (n_1) \) to denser \( (n_2) \): Here \( n_2 > n_1 \), so \( \frac{n_2}{n_1} > 1 \). This implies \( \frac{\sin i}{\sin r} > 1 \), which means \( \sin i > \sin r \) and \( i > r \). Thus, the ray bends towards the normal.
2. From denser \( (n_1) \) to rarer \( (n_2) \): Here \( n_2 < n_1 \), so \( \frac{n_2}{n_1} < 1 \). This implies \( \frac{\sin i}{\sin r} < 1 \), which means \( \sin i < \sin r \) and \( i < r \). Thus, the ray bends away from the normal.

Question. A concave lens has focal length of 15 cm. At what distance should be object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification of the image.
Answer: For a concave lens, \( f = -15 \text{ cm} \) and the image is virtual, so \( v = -10 \text{ cm} \).
Using lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{-10} - \frac{1}{u} = \frac{1}{-15} \)

\( \implies \) \( -\frac{1}{u} = -\frac{1}{15} + \frac{1}{10} = \frac{-2 + 3}{30} = \frac{1}{30} \)

\( \implies \) \( u = -30 \text{ cm} \).
The object should be placed 30 cm from the lens.
Magnification \( m = \frac{v}{u} = \frac{-10}{-30} = +0.33 \).
The image is virtual, erect, and diminished.