Reflection of Light at Curved Surfaces — Practice

Question 1

If an object is placed at C on the principal axis in front of a concave mirror, the position of the image is

Accepted Answer

at the centre of curvature

Question 2

All the distances related to spherical mirrors will be measured from

Accepted Answer

pole

Question 3

The size of the image formed by a convex mirror is always

Accepted Answer

diminished than object size

Question 4

Where will the image be formed when an object is placed on the principal axis of a concave mirror at a point between the focus and the centre of curvature?

Accepted Answer

The image is formed beyond the centre of curvature. It is enlarged, inverted and real, so it can be caught on a screen.

Question 5

Distinguish between real and virtual images.

Accepted Answer

A real image is formed by the actual intersection of reflected rays and can be caught on a screen. A virtual image is formed where the reflected rays only appear to meet, found by extending them backwards; it cannot be caught on a screen although it is visible when you look into the mirror.

Question 6

Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.

Accepted Answer

R = 8 cm so f = R/2 = 4 cm. Applying the sign convention, f = -4 cm and u = -10 cm. From 1/f = 1/v + 1/u: 1/v = 1/(-4) - 1/(-10) = -1/4 + 1/10 = -3/20. So v = -20/3 = -6.67 cm. The image forms about 6.67 cm in front of the mirror and is real.

Question 7

How do you get a virtual image using a concave mirror?

Accepted Answer

Place the object between the mirror and the focal point. The reflected rays then diverge and never meet, so no image appears on a screen. Extending them backwards gives a virtual image behind the mirror that is erect and enlarged.

Question 8

A convex mirror with a radius of curvature of 3 m is used as a rear-view mirror for a vehicle. If a bus is located 5 m from this mirror, find the position, nature and size of the image.

Accepted Answer

For a convex mirror f = R/2 = 1.5 m, positive by the sign convention since F lies behind the mirror, and u = -5 m. From 1/f = 1/v + 1/u: 1/v = 1/1.5 - 1/(-5) = 1/1.5 + 1/5 = 0.6667 + 0.2 = 0.8667, so v = +1.15 m approximately. The positive v places the image 1.15 m behind the mirror, so it is virtual. Magnification m = -v/u = -(1.15)/(-5) = +0.23, so the image is erect and about 0.23 times the size of the bus — strongly diminished.

Question 9

To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.

Accepted Answer

Place the object at the centre of curvature C. Every ray from the tip travelling towards C strikes the mirror along the normal and retraces its own path, so the reflected rays reconverge at the object's own position. The table confirms this row: object at C gives an image at C, same size, inverted and real.

Question 10

The magnification produced by a mirror is +1. What does it mean?

Accepted Answer

A magnification of +1 means the image is the same size as the object and erect, since the sign is positive. Because m = -v/u, m = +1 requires v = -u, which is the plane-mirror case: the image is as far behind the mirror as the object is in front, virtual and of the same size.

Practice — Reflection of Light at Curved Surfaces

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