Hold a concave mirror in your hand and direct its reflecting surface towards the sun. Direct the light reflected by the mirror on to a white card-board held close to the mirror. Move the card-board back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the sun on the sheet of paper; which is also termed as 'Principal Focus' of the concave mirror.

(a) List two applications of concave mirror.

(b) If the distance between the mirror and the principal focus is 15 cm, find the radius of curvature of the mirror.

(c) Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror.

OR

(c) An object 10 cm in size is placed at 100 cm in front of a concave mirror. If its image is formed at the same point where the object is located, find :

(i) focal length of the mirror, and

(ii) magnification of the image formed with sign as per Cartesian sign convention.

(a) Two applications of concave mirror are :

1. As a reflector — They are used in torches and vehicle headlights to get a powerful parallel beam of light.
2. As a shaving mirror — When a concave mirror is held near the face (such that the face is between pole and focus of the mirror), it gives an upright and magnified image. Hence even tiny hair can be seen.

(b) As we know, R = 2f where, R is the radius of curvature and f is the focal length.

Given focal length (f) = 15 cm

So, substituting we get,

R = 2 x 15 = 30 cm

(c) Ray diagram showing the type of image formed when an object is placed between pole and focus of a concave mirror is given below:

OR

(c) Given,

Object distance (u) = -100 cm

Image distance (v) = -100 cm

Focal length (f) = ?

[f in case of a concave mirror is negative.]

Using formula,

$\dfrac{1}{\text{f}}$ = $\dfrac{1}{\text{v}}$ + $\dfrac{1}{\text{u}}$

We get,

$\dfrac{1}{\text{f}}$ = $\dfrac{1}{\text{-100}}$ + $\dfrac{1}{\text{-100}}$

$\dfrac{1}{\text{f}}$ = $\dfrac{2}{\text{-100}}$ = $\dfrac{1}{\text{-50}}$

So, f = -50 cm

Hence, focal length = -50 cm.

(ii) Size of object (h_o) = 10 cm

magnification = ?

From formula,

m = - $\dfrac{\text{v}}{\text{u}}$

Substituting we get,

m = - $\dfrac{-100}{-100}$ = -1

Hence, magnification of the object = -1 i.e., inverted image is formed of same size as object.