At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side. What will be the magnification produced in this case?

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

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Updated on: 10-Oct-2022

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