Question

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 cmA 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(d)-60cm

Answer 1

Answer:

H0=10mm hi=5mm v=−30cm

Here  given ,

Now we know

m=−v u =hi ho ⇒u=−vho hi =−30×10 5 =−60 cm  Now using mirror formula  1 f =1 v +1 u =1 −30 +1 −60 =−1 20 Hence f=−20 cm

Answer 2

(b)$\huge\mathcal\purple{-20\:\:cm}$

★$\large\bf\red{Explanation✓}$

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=$\large\bf\red{ -v/u}$

→i/o=-V/u

→0.5/1=-30/u

→U = -60cm

→Focal length is given by 1/f = $\large\bf\red{-1/v + 1/u}$

→1/f=1/-30 + 1/60

= -2-1/60

=-3/60

F= -20cm