A concave lens has a focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

Given:

Focal length of the concave lens, $f$ = $-$20 cm

Image distance from the concave lens, $v$ = $-$15 cm              (image formed by a concave lens is on the left side of the lens)

Height of the object, $h$ = 5 cm

To find: Height of the image, $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the given values in the formula, we get-

$\frac {1}{(-15)}-\frac {1}{u}=\frac {1}{(-20)}$

$-\frac {1}{15}-\frac {1}{u}=-\frac {1}{20}$

$\frac {1}{20}-\frac {1}{15}=\frac {1}{u}$

$\frac {1}{u}=\frac {3-4}{60}$

$\frac {1}{u}=-\frac {1}{60}$

$u=-60cm$

Thus, the object is at a distance of 60 cm from the concave lens, and the negative sign implies that it is on the left side of it.

Now, 

Fro magnification we know that-

$m=\frac {v}{u}=\frac {h'}{h}$

Substituting the given values in the formula, we get-

$\frac {-15}{-60}=\frac {h'}{5}$

$\frac {1}{4}=\frac {h'}{5}$

$4\times {h'}=5$

$h'=\frac {5}{4}$

$h'=+1.25cm$

Thus, the size of the image is 1.25cm.

Tutorialspoint

Updated on: 10-Oct-2022

9K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started