Question

if a diverging lens is to be used to form an image which is one fourth of the size of the object where must the object be placed the diverging lens is to be used to form an image which is one fourth of the size of the object where must the object be placed?
a) 3f
b) 4f
c)2f
d)f ​

Answer 1

Answer:

4f

Explanation:

when the image is 1/4 to that real object the the image is 4f

Answer 2

Given : Diverging lens is to be used to form an image which is one fourth           of the size of the object.

To Find : Object placement distance

Solution :

As diverging lens forms an image

Then according to magnification formulae with relation is

            $\[\frac{{{h_i}}}{{{h_0}}} = - \frac{v}{u}\]$

putting value in equation,

        $\[\frac{v}{u} = - \frac{1}{4}\]$

Thus , v = $\[ - \frac{u}{4}\]$

putting value in lens formulae,

  $\[\left( {\frac{1}{v} + \frac{1}{u}} \right) = \frac{1}{f}\]$

  $\[\left( { - \frac{4}{u} + \frac{1}{u}} \right) = \frac{1}{f}\]$

  $\[ - \frac{3}{u} = \frac{1}{f}\]$

    u = -3f = $3|f|.$

Hence, object should be placed at 3f.