Question

Rays from the Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that the size of its image is equal to the size of the object?
A) 15 cm in front of the mirror
B) 30 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 1

Hint: A focus is a point at which rays of light, heat, or other radiation meet after being reflected or refracted. The focal point is the point in the space at which light incident towards the mirror and traveling parallel to the principal axis will meet after the reflection.

Complete step by step solution:

We know that from the surface of the earth, the Sun is at an infinite distance. So light rays from sun rays after the reflection from the concave mirror will converge at the focus.
From the above ray diagram it is clear that in the case of a concave mirror, the image size becomes equals to the object size when the object distance equals the radius if the curvature $ = $twice the focal length.
In the problem, it is given that the sun rays converge at 15 cm. So, the focal length of the mirror $ = $f $ = $ 15 cm.
In the case of a concave mirror, the size of the image and object will be the same if the object is placed at 2f.
$\therefore $ In this case, the object must be placed at 2f.
So substituting the value of f $ = $ 15 cm, we get
Object distance $ = 2 \times f = 2 \times 15 = 30$ cm

Hence the correct option for the problem is B.

Note: 1) A focus is also called an image point.
2) The principal focus is defined as the point where a beam parallel to the principal axis appears to diverge or converge from a point on the principal axis after passing through the lens.
3) A real image is formed by the collection of focus points made by the converging rays.
4) A virtual image is formed by the collection of focus points made by the extensions of diverging rays.