CBSE Class 10 Science Light Reflection and Refraction Worksheet Set 06

Answer: (b) virtual and erect. A plane mirror forms an erect virtual image behind the mirror.

Teacher's Note:
1. A plane mirror always forms a virtual and erect image of the same size as the object.
2. The image is also laterally inverted.

Answer: (c) real. Only real images can be received on a screen.

Teacher's Note:
1. Real images are formed by the actual meeting of light rays and can be caught on a screen.
2. Virtual images cannot be obtained on a screen.

Answer: (a) m = 1. The image in a plane mirror is as large as the object.

Teacher's Note:
1. The image has the same size as the object.
2. The magnification has no unit.

Answer: (a) 90°. By the law of reflection, the angle of reflection equals the angle of incidence.

Teacher's Note:
1. The angle of reflection is always equal to the angle of incidence.
2. A ray at 90° to the normal moves along the surface of the mirror.

Answer: (b) R = 2f. The principal focus lies at half the radius of curvature.

Teacher's Note:
1. The focus of a spherical mirror lies midway between the pole and the centre of curvature.
2. Hence the focal length is half the radius of curvature.

Answer: (c) P = 1/f. Power is the reciprocal of the focal length in metres.

Teacher's Note:
1. The SI unit of power is dioptre (D).
2. A lens of focal length 1 metre has a power of 1 D.

Answer: One dioptre is the power of a lens whose focal length is 1 metre.
The relation is P = 1/f, where f is the focal length in metres.

Teacher's Note:
1. Power measures the ability of a lens to converge or diverge light rays.
2. The power of a convex lens is positive and that of a concave lens is negative.

Answer: A convex mirror always forms a virtual, erect and diminished image between the pole and the focus behind the mirror.
Since the image distance v is positive and smaller in magnitude than the object distance, the magnification m = −v/u is positive and less than 1.

Teacher's Note:
1. The image is always smaller than the object in a convex mirror.
2. Virtual and erect images have positive magnification.

Answer: f = 32 cm = 0.32 m
P = 1/f = 1/0.32 = +3.125 D (about +3.1 D).

Teacher's Note:
1. The focal length must be converted into metres before using the formula.
2. The positive sign shows that the lens is a converging (convex) lens.

Answer: (a) Uses of a convex lens: magnifying glass or microscope; spectacles for correcting hypermetropia (also cameras).
(b) Uses of a convex mirror: rear-view mirror in vehicles; reflector in street lamps; security mirrors in shops.

Teacher's Note:
1. The uses follow from the properties of the images each one forms.
2. Give two correct uses for each part.

Answer: A convex mirror always gives an erect image.
It also has a wider field of view because it curves outwards, so the driver can see a larger area of traffic behind, though the image is diminished.

Teacher's Note:
1. The image is always erect, so the driver is not confused.
2. Its wide field of view is the main advantage.

Answer: The refractive index of a medium is the ratio of the speed of light in vacuum (or air) to the speed of light in that medium. It can also be given by sin i / sin r for light passing from air into the medium.
It depends on the nature of the medium, the colour (wavelength) of the light and the temperature.

Teacher's Note:
1. A medium of larger refractive index is optically denser.
2. Light travels more slowly in a medium of larger refractive index.

Answer: n(glass with respect to water) = n(glass) / n(water)
= (3/2) ÷ (4/3) = (3/2) × (3/4) = 9/8 = 1.125

Teacher's Note:
1. The ratio of the absolute refractive indices gives the relative refractive index.
2. The glass is optically denser than water, so the value is more than 1.

Answer: (a) There is no change in direction (no bending) when the light ray falls normally on the surface of separation, and when the two media have the same refractive index.
(b) Optical density tells us how a ray bends: it bends towards the normal on going into an optically denser medium and away from the normal on going into a rarer medium.

Teacher's Note:
1. Optical density is different from mass density.
2. A medium of higher refractive index is optically denser.

Answer: The refractive index of medium 2 with respect to medium 1 is n21 = v1/v2, where v1 and v2 are the speeds of light in the media 1 and 2.
The refractive index of medium 1 with respect to medium 2 is n12 = v2/v1.
Therefore n21 × n12 = (v1/v2) × (v2/v1) = 1. Hence proved.

Teacher's Note:
1. The relative refractive index is the ratio of the speeds of light in the two media.
2. Hence n12 is the reciprocal of n21.

Answer: For a concave lens, f = −15 cm and the image is virtual, on the same side as the object, so v = −10 cm.
1/v − 1/u = 1/f gives 1/u = 1/v − 1/f = −1/10 + 1/15 = −1/30.
Therefore u = −30 cm, so the object should be placed 30 cm from the lens.
Magnification m = v/u = (−10)/(−30) = +1/3 (about +0.33). The image is virtual, erect and diminished.

Teacher's Note:
1. A concave lens always forms a virtual image on the same side as the object.
2. The sign convention must be applied carefully to each quantity.

Answer: u = −10 cm and f = −15 cm.
1/v + 1/u = 1/f gives 1/v = −1/15 + 1/10 = 1/30, so v = +30 cm.
m = −v/u = −30/(−10) = +3.
The image is formed 30 cm behind the mirror; it is virtual, erect and three times the size of the object.

Teacher's Note:
1. When the object is between the pole and the focus of a concave mirror, the image is virtual and magnified.
2. This is how a shaving mirror works.

Answer: n = c / v, so v = c / n = (3 × 10⁸) / 1.5 = 2 × 10⁸ m/s.

Teacher's Note:
1. Light travels slower in glass than in vacuum.
2. The speed of light in a medium is found by dividing c by the refractive index.

Answer: f = −25 cm = −0.25 m
P = 1/f = 1/(−0.25) = −4 D.

Teacher's Note:
1. The negative sign of power shows a diverging lens.
2. The focal length of a concave lens is always taken as negative.