Question

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Answer 1

here is ur answer.......................

Answer 2

Answer:

-5 cm

Explanation:

According to this question

Given that,

u = -27 cm, h = 2.5 cm,

R = 2f = 36

f = $\frac{-36}{2}$

f = -18       (with sign convention)

By  mirror formula,

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

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

$\frac{1}{v}$ = $\frac{1}{-18} - \frac{1}{27}$

$\frac{1}{v}$ = $\frac{-3 + 2}{54}$

$\frac{1}{v}$ = $\frac{-1}{54}$

$v$ = $-54$

Thus, the image is formed in front of mirror at a distance 54 cm from the mirror.

Therefore, the screen must be placed at a distance of 54 cm from the mirror.

Size of the image $h'$ = $\frac{-v}{u}\times h$

h' = $\frac{-(-54)}{-27}\times 2.5$

h' = 5 cm