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