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? *​

Answer 1

Explanation:

Given : u=−0.5 m

Convex mirror forms erect image of real object.

Using,

h

o

h

i

=−

u

v

Or

4

1

=−

(−0.5)

v

⟹ v=0.125 m

Using mirror formula,

v

1

+

u

1

=

f

1

∴

0.125

1

+

−0.5

1

=

f

1

⟹ f=0.16 m

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,

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

putting value in lens formulae,

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

$\[ - \frac{3}{u} = \frac{1}{f}\]$

u = -3f = $\[\left| {3f} \right|\]$

Hence, object should be placed at 3f.