1. A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is (a) – 30 cm (b) – 20 cm (c) – 40 cm – 60 cm

• Soln:

Answer is (b) – 20 cm

Explanation:

Here, size of object = O = + 10.0 mm = + 1.0 cm (as, 1 cm = 10 mm) Size of Image size = I = 5.0 mm = 0.5 cm

Image distance, v = − 30 cm (as image is real) Let, object distance = u

Focal length, f =?

Magnification m= I (Size of image)

O(Size of image)

Magnification is given by m=^−𝑣

𝑢

𝐼 = −𝑣

𝑂 𝑢

0.5=−30

1 𝑢

U= -60cm

Focal length is given by ¹= ¹+¹

𝑓 𝑣 𝑢

1= 1 + 1

𝑓 −30 −60

=−2−1

60

=−3

60

F= -20cm