Question

The image of an object is formed on a screen at 3/2 of the distance of the object using a
concave mirror of focal length 9 cm. Find the object distance, image distance and
magnification​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

It is given that :-

Image distance = v = 3/2u = ?

Object distance = u

Focal length = 9cm

Magnification = ?

______________________________________

Now according to mirror formula :

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

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

After inputting the known values in this equation we get :

$9 = \frac{ \frac{3u}{2} \times u }{ \frac{3u}{2} + u }$

$9 = \frac{3 {u}^{2} }{2} \div \frac{5u}{2}$

$9 = \frac{3u}{5}$

$3u = 45 \\ u = \frac{45}{3} \\ \\ u = 15cm$

Thus we got the image distance is 15cm.

Now we know v = 3/2u

Thus the image distance is :

$\frac{3}{2} \times 15$

$v = 22.5cm$

Thus the image distance is 22.5cm.

And now magnification is :

$m = \frac{ - v}{u}$

After inputting the values :

$m = \frac{ - 22.5}{15}$

$m = - 1.5cm$

Thus the magnification of image is -1.5 cm.

______________________________________

BY SCIVIBHANSHU

THANK YOU

STAY CURIOUS