At what distance should an object be placed from convex lens of focal length 18 cm to obtain an image at 36 cm from it?

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

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

Text Solution

Solution : Here, `u = ?, f = 18 cm, v = +- 36 cm`, <br> `m = ?` <br> From `(1)/(v) - (1)/(u) = (1)/(f)`, `-(1)/(u) = (1)/(f) -(1)/(u) = (1)/(18) - (1)/(+- 36) = (2-(+-1))/(36)` <br> `:. u =-36 cm` and `u = -12 cm`. <br> `{:("When u"= -36 cm),(and v = 36 cm):}],m = (v)/(u) = (+36)/(-36) = -1`, Image is real and inverted. <br> `{:("When u" = -12 cm),( and v = -36 cm):}], m = (v)/(u) = (-36)/(-12) = +3`, Image is virtual and erect.