At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at?

AcademicPhysicsNCERTClass 10

Given,

The focal length of a convex lens, f = 18 cm.

Image distance, v = 24 cm

Object distance, u = ?

To find- Magnification

Solution:

By using lens formula-

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$

where, v = image distance, u = object distance, and f = focal length

Substituting the values of f, v and u we get,

$\frac{1}{24}-\frac{1}{u}=\frac{1}{18}\phantom{\rule{0ex}{0ex}}$

$\frac{1}{24}-\frac{1}{18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$

$\frac{18-24}{24\times 18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$

$\frac{6}{24\times 18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$

$\frac{1}{4\times 18}=\frac{1}{u}\phantom{\rule{0ex}{0ex}}$

$u=-72cm\phantom{\rule{0ex}{0ex}}$

So, the object distance is -72cm.

The object should be placed at a distance of -72 cm from the lens.

Now, the equation for finding magnification of a lens can be given as-

$m=\frac{v}{u}\phantom{\rule{0ex}{0ex}}$

Substituting the values in magnification formula we get-

$m=\frac{24}{-72}\phantom{\rule{0ex}{0ex}}$

$m=-\frac{1}{3}\phantom{\rule{0ex}{0ex}}$

Hence, the magnification produced will be $m=-\frac{1}{3}\phantom{\rule{0ex}{0ex}}$

Updated on 10-Oct-2022 10:29:43

Text Solution

Solution : Here, `u=?, f= 18cm, v = 24 cm, m=?` <br> from `(1)/(v) -1/u =(1)/(f), - 1/u = (1)/(f) - (1)/(v) =1/18 - 1/24 = (4-3)/(72)=1/72` <br> `u =- 72 cm`. <br> As `m = v/u = (-72)/(24)= -3` <br> Negative sign of m indicates that the image would be inverted.