At what distance should an object be placed from a convex lens of focal length 18 cm to obtain a image at 24 cm from it on the other side?

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

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:F length, f = 18 cmI distance,v = 24 cmPng these values in lens formula, we get:1- 1/u = 1/fo/u = 1/v-1/fo1/u = 1/24-1/18 =-(1/72)o =-72 cmT the object should be placed at a distance of 72 cm from the lens.N

Mfication, m = v /u = 24/ (-72) =-(1/3)

Concept: Linear Magnification (M) Due to Spherical Mirrors

  Is there an error in this question or solution?

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