Question

4.
A linear object of 10 cm length lies along the axis of a concave mirror of radius of curva
30 cm, the nearer end lying 18 cm from the mirror. The magnification produced is:
A) 5.77
B) 6.77
C) 7.77
D) 8.77​

Answer 1

Answer:

I think a=5.77 ........

Answer 2

Answer:

Magnification produced is -5

Explanation:

u = 18 cm

r = 30 cm

f = r/2 = 15 cm

From relation,

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

For concave mirror,

u is negative

v is negative

f is negative

Thus,

$\frac{1}{-18} + \frac{1}{v} = \frac{1}{-15}$

$-\frac{1}{18} + \frac{1}{v} = -\frac{1}{15}$

$\frac{1}{v} = -\frac{1}{15} + \frac{1}{18}$

$\frac{1}{v} = \frac{-6 + 5}{90}$

$\frac{1}{v} = \frac{-1}{90}$

v = -90 cm

Magnification = -v/u

M = $-\frac{(-90)}{-18}$

M = - 5