CBSE Class 10 Science Reflection of Light and Spherical Mirrors Worksheet Set 04

Answer: Light is a form of energy that produces the sensation of vision and travels as an electromagnetic wave. Its speed in vacuum is 3 × 10⁸ m/s.

Teacher's Note:
1. Light needs no material medium for travel.
2. The value 3 × 10⁸ m/s is the speed in vacuum only.

Answer: The pole (P) is the centre of the reflecting surface of the mirror. The centre of curvature (C) is the centre of the hollow sphere of which the mirror is a part.

Teacher's Note:
1. The pole lies on the mirror itself, while C lies away from it.
2. The distance PC is the radius of curvature.

Answer: A spherical mirror is a mirror whose reflecting surface is a part of a hollow sphere. Its types are concave (reflecting surface curved inwards) and convex (reflecting surface curved outwards).

Teacher's Note:
1. A concave mirror is also called a converging mirror.
2. A convex mirror is also called a diverging mirror.

Answer:

1. The angle of incidence is equal to the angle of reflection.

2. The incident ray, the reflected ray and the normal to the mirror at the point of incidence all lie in the same plane.

Teacher's Note:
1. Angles are always measured from the normal, not from the surface.
2. The laws apply at every point of a curved mirror.

Answer:

1. A concave mirror curves inwards; a convex mirror curves outwards.

2. A concave mirror converges parallel rays to a real focus; a convex mirror diverges them so that they appear to come from a virtual focus.

3. A concave mirror can form real or virtual images; a convex mirror always forms a virtual, erect and diminished image.

Teacher's Note:
1. Think of the reflecting surface as the inside or outside of a bowl.
2. Compare the nature of the focus and of the images formed.

Answer: 0°. At normal incidence the ray goes straight through the slab without bending, so there is no lateral shift.

Teacher's Note:
1. Lateral shift depends on the bending of the ray inside the slab.
2. No bending occurs along the normal.

Answer: Medium 1 is optically denser. Its angle of refraction (20°) is smaller, so the ray bends more towards the normal, which means light travels slower in it and its refractive index (sin 45°/sin 20°) is larger than that of medium 2 (sin 45°/sin 30°).

Teacher's Note:
1. A ray bends towards the normal on entering a denser medium.
2. The smaller the angle of refraction, the higher the refractive index.

Answer: Medium r. The refractive index is n = sin i / sin r, so the smallest angle of refraction (15°) gives the largest n. Since v = c/n, the speed is least in r.

Teacher's Note:
1. The medium with the greatest bending towards the normal is optically densest.
2. Speed of light is inversely proportional to refractive index.

Answer:

1. The image is virtual and erect.

2. It is of the same size as the object.

3. It is as far behind the mirror as the object is in front.

4. It is laterally inverted (left and right are interchanged).

Teacher's Note:
1. Virtual means the image cannot be obtained on a screen.
2. Lateral inversion is why the word AMBULANCE is written reversed on vehicles.

Answer: +1. Since h_i = h_o, m = h_i/h_o = +1; the plus sign shows that the image is erect.

Teacher's Note:
1. Magnification is the ratio of image height to object height.
2. A positive sign means an erect image.

Answer: (c) R = 2f. The focus lies midway between the pole and the centre of curvature, so f = R/2.

Teacher's Note:
1. Focal length is half the radius of curvature.
2. This holds for both concave and convex mirrors.

Answer: (c) along the same path. A ray through C strikes the mirror normally, so it retraces its own path.

Teacher's Note:
1. A ray along the normal has zero angle of incidence.
2. It therefore returns along the same line.

Answer: (b) virtual, erect and diminished. A convex mirror forms only virtual, erect and diminished images between P and F behind the mirror.

Teacher's Note:
1. This property makes the field of view wide.
2. It does not depend on the position of the object.

Answer: Using the Cartesian sign convention: u = −30 cm, f = −15 cm.
\( \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{15} + \dfrac{1}{30} = -\dfrac{1}{30} \)
So v = −30 cm. Magnification m = −v/u = −1.
The image is formed 30 cm in front of the mirror, real, inverted and of the same size as the object (the object is at C).

Teacher's Note:
1. Put u and f negative for a concave mirror with the object in front.
2. An object at the centre of curvature gives an image of equal size at the same place.

Answer: f = +20 cm, u = −40 cm.
\( \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{40} = \dfrac{3}{40} \), so v = +13.3 cm (behind the mirror).
m = −v/u = 13.3/40 = +1/3, so h_i = 6 × 1/3 = +2 cm.
The image is virtual, erect and diminished.

Teacher's Note:
1. Focal length of a convex mirror is positive.
2. A positive magnification below 1 means an erect, smaller image.

Answer: 15 cm, since f = R/2 = 30/2. It is taken as +15 cm for a convex mirror.

Teacher's Note:
1. Use f = R/2.
2. The sign of f is positive for convex mirrors.

Answer: A convex mirror always forms an erect, diminished image and has a wider field of view than a plane mirror. The driver can therefore see a larger area of traffic behind the vehicle.

Teacher's Note:
1. Wide field of view is the key advantage.
2. The image appears smaller, so the driver must judge distances carefully.

Answer:

1. Used as a shaving mirror or dentist's mirror because it forms an enlarged, erect image when the object is within the focal length.

2. Used in torches, headlights and solar furnaces to get a strong parallel beam or to concentrate sunlight.

Teacher's Note:
1. A source at the focus gives a parallel beam after reflection.
2. A concave mirror concentrates parallel rays at its focus.