CBSE Class 10 Science HOTs Reflection and Refraction Set 06

Question. State Snell’s law of refraction of light. Write an expression for the absolute refractive index of a medium in terms of speed of light.
Answer: Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media and for light of a given colour. The expression for the absolute refractive index (\( n \)) of a medium is:
\( n = \frac{c}{v} \), where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the medium.

Question. Light enters from air to glass having refractive index 1.50. Calculate the speed of light in the glass. Given : The speed of light in vacuum is \( 3 \times 10^8 \) m/s.
Answer: Speed of light in vacuum (\( c \)) = \( 3 \times 10^8 \) m/s
Refractive index of glass (\( n \)) = 1.50
Speed of light in glass (\( v \)) = \( \frac{c}{n} \)

\( \implies v = \frac{3 \times 10^8}{1.50} \)

\( \implies v = 2 \times 10^8 \text{ m/s} \).

Question. An object is placed at a distance of 15 cm from a convex lens of focal length 20 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.
Answer: For a convex lens, since the object distance (15 cm) is less than the focal length (20 cm), the object is between the focus and the optical centre. The characteristics of the image are:
1. The image is virtual.
2. The image is erect.
3. The image is magnified (enlarged).
4. The image is formed on the same side of the lens as the object.

Question. What is meant by power of a lens? What does its sign (+ve or –ve) indicate? State its S.I. unit related to focal length of a lens.
Answer: Power of a lens is the measure of its ability to converge or diverge light rays. It is the reciprocal of its focal length in meters. A +ve sign indicates a converging (convex) lens, while a –ve sign indicates a diverging (concave) lens. The S.I. unit is Dioptre (D).

Question. The refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. If speed of light in glass is \( 2 \times 10^8 \) m/s, find the speed of light in water.
Answer: Refractive index of glass (\( n_g \)) = 3/2
Refractive index of water (\( n_w \)) = 4/3
Speed of light in air (\( c \)) = \( n_g \times v_g = \frac{3}{2} \times 2 \times 10^8 = 3 \times 10^8 \text{ m/s} \).
Speed of light in water (\( v_w \)) = \( \frac{c}{n_w} = \frac{3 \times 10^8}{4/3} \)

\( \implies v_w = 2.25 \times 10^8 \text{ m/s} \).

Question. The absolute refractive indices of glass and water are 4/3 and 3/2 respectively. If the speed of light in glass is \( 2 \times 10^8 \) m/s, calculate the speed of light in (i) vacuum, (ii) water.
Answer: (i) Speed of light in vacuum (\( c \)) = \( n_g \times v_g = \frac{4}{3} \times 2 \times 10^8 = 2.67 \times 10^8 \text{ m/s} \).
(ii) Speed of light in water (\( v_w \)) = \( \frac{c}{n_w} = \frac{2.67 \times 10^8}{3/2} \)

\( \implies v_w \approx 1.78 \times 10^8 \text{ m/s} \).

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: \( h_o = 10 \) cm, \( u = -25 \) cm, \( f = +15 \) cm.
Lens formula: \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{15} - \frac{1}{25} = \frac{5-3}{75} = \frac{2}{75} \)

\( \implies v = 37.5 \text{ cm} \).
Height of image (\( h_i \)):
\( m = \frac{v}{u} = \frac{37.5}{-25} = -1.5 \)

\( \implies h_i = m \times h_o = -1.5 \times 10 = -15 \text{ cm} \).
Image-distance is 37.5 cm and image height is 15 cm (inverted).

Question. Define power of a lens. The focal length of a lens is –10 cm. Write the nature of the lens and find its power. If an object is placed at a distance of 20 cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case?
Answer: Power of a lens is the reciprocal of its focal length in meters. Since \( f = -10 \) cm, the lens is concave. Power \( P = \frac{1}{f} = \frac{1}{-0.1} = -10 \text{ D} \). For a concave lens, the image is always virtual and erect, so the magnification will always be positive (+ve).

Question. The refractive index of a medium ‘x’ with respect to a medium ‘y’ is 2/3 and the refractive index of medium ‘y’ with respect to medium ‘z’ is 4/3. Find the refractive index of medium ‘z’ with respect to medium ‘x’. If the speed of light in medium ‘x’ is \( 3 \times 10^8 \) m s^–1, calculate the speed of light in medium ‘y’.
Answer: \( n_{xy} = 2/3 \), \( n_{yz} = 4/3 \).
\( n_{zx} = \frac{1}{n_{yz}} \times \frac{1}{n_{xy}} = \frac{3}{4} \times \frac{3}{2} = \frac{9}{8} \).
Speed in 'y': \( n_{xy} = \frac{v_y}{v_x} \implies \frac{2}{3} = \frac{v_y}{3 \times 10^8} \)

\( \implies v_y = 2 \times 10^8 \text{ m/s} \).

Question. How far should an object be placed from a convex lens of focal length 20 cm to obtain its real image at a distance of 30 cm from the lens ? Determine the height of the image if the object is 4 cm tall.
Answer: \( f = +20 \) cm, \( v = +30 \) cm, \( h_o = 4 \) cm.
Lens formula: \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{30} - \frac{1}{20} = -\frac{1}{60} \implies u = -60 \text{ cm} \).
\( m = \frac{v}{u} = \frac{30}{-60} = -0.5 \). \( h_i = m \times h_o = -2 \text{ cm} \). Height of image is 2 cm (inverted).

Question. A real image 2/3^rd of the size of an 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: \( m = -2/3 \), \( u = -12 \) cm.
\( m = \frac{v}{u} \implies -\frac{2}{3} = \frac{v}{-12} \implies v = +8 \text{ cm} \).
Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{8} + \frac{1}{12} = \frac{5}{24} \)

\( \implies f = 4.8 \text{ cm} \).

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 vaccum.
Answer: Laws of refraction: 1. The incident ray, refracted ray, and normal at the point of incidence lie in the same plane. 2. Snell's Law states \( \frac{\sin i}{\sin r} = \text{constant} \). Absolute refractive index is the ratio of speed of light in vacuum to speed of light in the medium: \( n = \frac{c}{v} \).

Question. What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of –20 cm. Write the nature and power of each lens.
Answer: Power is the reciprocal of focal length in meters. Unit is Dioptre (D).
Lens 1 (\( f = 40 \) cm): Power \( = \frac{1}{0.4} = +2.5 \text{ D} \), Nature: Convex.
Lens 2 (\( f = -20 \) cm): Power \( = \frac{1}{-0.2} = -5 \text{ D} \), Nature: Concave.

Question. State the laws of refraction of light. If the speed of light in vacuum is \( 3 \times 10^8 \) m/s, find the absolute refractive index of a medium in which light travels with a speed of \( 1.4 \times 10^8 \) m/s.
Answer: Absolute refractive index \( n = \frac{c}{v} = \frac{3 \times 10^8}{1.4 \times 10^8} \approx 2.14 \).

Question. State the laws of refraction of light. If the speed of light in vacuum is \( 3 \times 10^8 \) m s^–1, find the speed of light in a medium of absolute refractive index 1.5.
Answer: Speed of light in medium \( v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s} \).

Question. Determine the speed of light in diamond if the refractive index of diamond with respect to vacuum is 2.42. Speed of light in vacuum is \( 3 \times 10^8 \) m/s.
Answer: Refractive index (\( n \)) \( = 2.42 \)
Speed of light in vacuum (\( c \)) \( = 3 \times 10^8 \text{ m/s} \)
Using formula \( n = \frac{c}{v} \)
\( 2.42 = \frac{3 \times 10^8}{v} \)

\( \implies \) \( v = \frac{3 \times 10^8}{2.42} \)

\( \implies \) \( v = 1.239 \times 10^8 \text{ m/s} \)

Question. Refractive indices of glass, water and carbon disulphide are 1.5, 1.33 and 1.62 respectively. If a ray of light is incident in these media at the same angle (say \( \theta \)), then write the increasing order of the angle of refraction in these media.
Answer: Higher the refractive index, more is the bending towards the normal, hence smaller is the angle of refraction. The refractive indices are: Water (1.33) < Glass (1.5) < Carbon disulphide (1.62).
Therefore, the increasing order of the angle of refraction is: Carbon disulphide < Glass < Water.

Question. The speed of light in glass is \( 2 \times 10^8 \) m/s and in water is \( 2.25 \times 10^8 \) m/s. (a) Which one of the two is optically denser and why? (b) A ray of light is incident normally at the water-glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass? Give reason.
Answer: (a) Glass is optically denser because the speed of light in glass (\( 2 \times 10^8 \text{ m/s} \)) is less than the speed of light in water (\( 2.25 \times 10^8 \text{ m/s} \)). (b) The ray of light will pass undeviated (straight) into the glass because when light is incident normally (\( \angle i = 0^\circ \)) on an interface, the angle of refraction is also zero (\( \angle r = 0^\circ \)).

Question. The absolute refractive indices of water and glass are 4/3 and 3/2 respectively. If the speed of light in glass is \( 2 \times 10^8 \) m/s, find the speed of light in (i) vacuum and (ii) water.
Answer: (i) Speed of light in vacuum (\( c \)) \( = n_g \times v_g = \frac{3}{2} \times 2 \times 10^8 = 3 \times 10^8 \text{ m/s} \).
(ii) Speed of light in water (\( v_w \)) \( = \frac{c}{n_w} = \frac{3 \times 10^8}{4/3} = \frac{3 \times 3}{4} \times 10^8 = 2.25 \times 10^8 \text{ m/s} \).

Many optical instruments consists of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses places in contact is given by the algebraic sum of the powers of the individual lenses \( P_1 \), \( P_2 \), \( P_3 \), .... as \( P = P_1 + P_2 + P_3 + \dots \). This is also termed as the simple additive property of the power of lens, widely used to design lens systems of cameras, microscopes and telescopes. These lens systems can have a combination of convex lenses and also concave lenses.

Question. What is the nature (convergent/divergent) of the combination of a convex lens of power +4 D and a concave lens of power –2 D?
Answer: Net Power \( P = P_1 + P_2 = +4 \text{ D} + (-2 \text{ D}) = +2 \text{ D} \).
Since the net power is positive, the nature of the combination is convergent.

Question. Calculate the focal length of a lens of power –2.5 D.
Answer: \( f = \frac{1}{P} = \frac{1}{-2.5} \text{ m} = -0.4 \text{ m} = -40 \text{ cm} \).

Question. How is a virtual image formed by a convex lens different from that formed by a concave lens? Under what conditions do a convex and a concave lens form virtual image?
Answer: A virtual image formed by a convex lens is always magnified, whereas a virtual image formed by a concave lens is always diminished. A convex lens forms a virtual image only when the object is placed between the optical centre and its principal focus. A concave lens forms a virtual image for all positions of the object.

Question. Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose.
(i) State the nature of the lens and reason for its use.
(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?
(iii) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image.
Answer: (i) Nature of lens: Convex lens. Reason: It forms a virtual, erect, and magnified image of the lines on the palm when held close. (ii) For a real and magnified image, the palmist should hold the lens such that the palm is between \( F \) and \( 2F \) of the lens. (iii) Given \( f = +10 \text{ cm} \), \( u = -5 \text{ cm} \).
Using lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{v} = \frac{1}{10} + \frac{1}{-5} = \frac{1-2}{10} = -\frac{1}{10} \)

\( \implies \) \( v = -10 \text{ cm} \).
Magnification \( m = \frac{v}{u} = \frac{-10}{-5} = 2 \).
The image is virtual, erect and twice the size of the object.

Question. An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.
Answer: \( u = -20 \text{ cm} \), \( f = -10 \text{ cm} \).
Using \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-10} + \frac{1}{-20} = -\frac{3}{20} \)

\( \implies \) \( v = -6.67 \text{ cm} \).

Question. The image of a candle flame placed at a distance of 30 cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 60 cm from the optical centre of the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 3 cm, find the height of its image.
Answer: Image on screen \( \implies \) real image \( \implies \) lens is convex. \( u = -30 \text{ cm} \), \( v = +60 \text{ cm} \).
\( \frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20} \)

\( \implies \) \( f = 20 \text{ cm} \).
\( m = \frac{v}{u} = \frac{60}{-30} = -2 \).
\( h_i = -2 \times 3 = -6 \text{ cm} \).

Question. (a) 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. (b) The absolute refractive indices of two media A and B are 2.0 and 1.5 respectively. If the speed of light in medium B is \( 2 \times 10^8 \) m/s. Calculate the speed of light in (i) vacuum (ii) medium A.
Answer: (b)(i) \( c = n_B \times v_B = 1.5 \times 2 \times 10^8 = 3 \times 10^8 \text{ m/s} \).
(ii) \( v_A = \frac{c}{n_A} = \frac{3 \times 10^8}{2.0} = 1.5 \times 10^8 \text{ m/s} \).

Question. (a) Explain the following terms related to spherical lenses: (i) optical centre (ii) centre of curvature (iii) principal axis (iv) aperture (v) principal focus (vi) 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: (b) \( f = +12 \text{ cm} \), \( v = +48 \text{ cm} \).
\( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{48} - \frac{1}{12} = \frac{1-4}{48} = -\frac{3}{48} = -\frac{1}{16} \)

\( \implies \) \( u = -16 \text{ cm} \).

Question. (ii) At what distance from a concave lens of focal length 20 cm, should a 6 cm tall object be placed so that it forms an image at 15 cm from the lens? Also determine the size of the image formed.
Answer: \( f = -20 \text{ cm} \), \( v = -15 \text{ cm} \).
Using lens formula: \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-15} + \frac{1}{20} = -\frac{1}{60} \)

\( \implies \) \( u = -60 \text{ cm} \).
Magnification \( m = \frac{v}{u} = \frac{-15}{-60} = 0.25 \).
Size \( h_i = 0.25 \times 6 = 1.5 \text{ cm} \).

Question. What is meant by power of a lens? Name and define its S.I. unit. One student uses a lens of focal length +50 cm and another of –50 cm. State the nature and find the power of each lens. Which of the two lenses will always give a virtual and diminished image irrespective of the position of the object?
Answer: Lens 1 (\( f = +50 \text{ cm} \)): nature: convex, \( P = \frac{1}{0.5} = +2 \text{ D} \). Lens 2 (\( f = -50 \text{ cm} \)): nature: concave, \( P = \frac{1}{-0.5} = -2 \text{ D} \). Concave lens always gives virtual and diminished image.

Questions

Spherical Mirrors

Question. Which of the following mirror is used by a dentist to examine a small cavity in a patient’s teeth?
(a) Convex mirror
(b) Plane mirror
(c) Concave mirror
(d) Any spherical mirror
Answer: (c) Concave mirror

Spherical Mirrors

Question. Rays from Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that size of its image is equal to the size of the object?
(a) 30 cm in front of the mirror
(b) 15 cm in front of the mirror
(c) Between 15 cm and 30 cm in front of the mirror
(d) More than 30 cm in front of the mirror
Answer: (a) 30 cm in front of the mirror

Question. The image formed by a concave mirror is observed to be real, inverted and larger than the object. Where is the object placed?
Answer: The object is placed between the centre of curvature (C) and the principal focus (F) of the concave mirror.

Question. Rohit wants to have an erect image of an object using a converging mirror of focal length 40 cm.
(a) Specify the range of distance where the object can be placed in front of the mirror. Justify.
(b) Draw a ray diagram to show image formation in this case.
(c) State one use of the mirror based on the above kind of image formation.
Answer: (a) The object should be placed at a distance between 0 cm and 40 cm from the mirror (i.e., between the pole and the focus). This is because a concave mirror forms a virtual and erect image only when the object is within its focal length.
(b) [A ray diagram should be drawn showing an object between the pole P and focus F of a concave mirror, with rays diverging and appearing to meet behind the mirror to form a magnified, erect image.]
(c) Such a mirror is used as a shaving mirror or a dentist's mirror to see a larger, erect image of the face or teeth.

Refraction of Light

Question. Which of the following can make a parallel beam of light when light from a point source is incident on it?
(a) Concave mirror as well as convex lens.
(b) Convex mirror as well as concave lens.
(c) Two plane mirrors placed at 90° to each others.
(d) Concave mirror as well as concave lens.
Answer: (a) Concave mirror as well as convex lens.

Question. Consider these indices of refraction: glass: 1.52; air: 1.0003; water: 1.333. Based on the refractive indices of three materials, arrange the speed of light through them in decreasing order.
(a) The speed of light in water > the speed of light in air > the speed of light in glass.
(b) The speed of light in glass > the speed of light in water > the speed of light in air.
(c) The speed of light in air > the speed of light in water > the speed of light in glass.
(d) The speed of light in glass > the speed of light in air > the speed of light in water.
Answer: (c) The speed of light in air > the speed of light in water > the speed of light in glass.

Question. If the power of a lens is –4.0 D, then it means that the lens is a
(a) concave lens of focal length –50 m
(b) convex lens of focal length +50 cm
(c) concave lens of focal length –25 cm
(d) convex lens of focal length –25 m
Answer: (c) concave lens of focal length –25 cm

Question. If the real image of a 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. The refractive index of flint glass is 1.65 and that for alcohol is 1.36 with respect to air. What is the refractive index of the flint glass with respect to alcohol ?
(a) 0.82
(b) 1.21
(c) 1.11
(d) 1.01
Answer: (b) 1.21

Question. Both a spherical mirror and a thin spherical lens have a focal length of (–) 15 cm. What type of mirror and lens are these?
Answer: Both are concave. A concave mirror and a concave lens have negative focal lengths.

Question. Name the part of a lens through which a ray of light passes without suffering any deviation.
Answer: The optical centre of the lens.

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.
(i) What is the magnification obtained by using the glass?
(ii) She keeps a book at a distance 10 cm from her eyes and tries to read. She is unable to read. What is the reason for this?
(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)(i) For a simple microscope (magnifying glass), magnification \( m = 1 + \frac{D}{f} \).
Given \( D = 25 \text{ cm} \), \( f = 5 \text{ cm} \).

\( \implies \) \( m = 1 + \frac{25}{5} = 1 + 5 = 6 \).
(ii) The least distance of distinct vision for a normal human eye is 25 cm. If an object is kept closer than this distance (e.g., at 10 cm), the eye muscles cannot strain enough to focus the image on the retina, resulting in a blurred image.
(b) Hari is unable to read because the book is kept at 10 cm, which is less than the normal near point (25 cm) of the eye. Objects must be at or beyond the near point for clear vision without excessive strain.