Question

An object is 2 m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?

Answer 1

$\textrm{Focal length}$ $f = \infty$  

The focal length of lens is infinity so it behaves as plane mirror.

Explanation:

Given that,

$\textrm{Distance of the object from the lens}$ $u = \textrm{-2 metre}$

$\textrm{Maginification of object}$ $m = \frac{1}{4}$

$\textrm{Maginification of object is given by}$

$m = \frac{v}{u}$

Where, $v = \textrm{distance of the image from lens}$

$\frac{1}{4} = \frac{v}{-2}$

$v =\textrm{-2 metre}$

$\textrm{the focal length (f) is given by}$

$\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$

$\frac{1}{f} = \frac{1}{-2}-\frac{1}{-2}$

$\frac{1}{f} = 0$

$f = \infty$  

Hence, the focal length of lens is infinity so it behaves as plane mirror.

Answer 2

Lens is Concave type

Explanation:

u = - 2m

m = $\frac {1}{4} = \frac {v}{u}$

or $\frac {1}{4} = \frac {v}{-2}$

v = $\frac {1}{-2}$ = - 0.5 m

Also,

$\frac {1}{v} - \frac {1}{u} = \frac {1}{f}$

= $\frac {1}{-0.5} - \frac {1}{-2} = \frac {1}{f}$

or

$\frac {1}{f} = - \frac {3}{2}$

f = $- \frac {2}{3}$

Which proves that lens is Concave!