Question

. An object, 4 cm in height, is placed at 25 cmfrom a concave mirror of focal length 15 cm.At what distance from the mirror should a screen be kept, so that a sharp image of the object can be obtained on the screen? Find
the nature and height of the image ​

Answer 1

Hola mate here is your answer!

Given:

ho=4cm

u=(-25)cm

Concave mirror

f=(-15)cm

To find:

v=?

hi=?

Nature

Solution:

Using mirror formula:

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

$\frac{1}{ - 15} = \frac{1}{v} - \frac{1}{25}$

$\frac{1}{v} = \frac{1}{25} - \frac{1}{15}$

$\frac{1}{v} = \frac{ - 10}{375}$

$v = - 37.5cm$

Now using Magnification formula:

$\frac{ - v}{u} = \frac{hi}{ho}$

$\frac{ 37.5}{ - 25} = \frac{hi}{4}$

$\frac{ - 3}{2} = \frac{hi}{4}$

$hi = 4 \times \frac{ - 3}{2}$

$hi = - 6cm$

Thus, The nature of image is

Real, inverted and Enlarged.

Hope it helps!