Q65 / 90 · JEE Main 2021 (31 Aug Shift 1) · Physics

An object is placed at the focus of concave lens having focal length $f .$ What is the magnification and distance of the image from the optical centre of the lens?

A
$\frac{1}{4}, \frac{f}{4}$
Correct answer
B
$\frac{1}{2}, \frac{f}{2}$
C
Very high, $\infty$
D
$1, \infty$

Solution

\frac{1}{F} = \frac{1}{v} - \frac{1}{u}

Put

u = - f

and

F = - f

\frac{1}{- f} = \frac{1}{v} - \frac{1}{- f}; \frac{- 1}{f} = \frac{1}{f} - \frac{1}{v}; \frac{- 2}{f} = \frac{1}{v} \Rightarrow v = - \frac{f}{2}

Magnification

= v / u = \frac{- f}{2 (- f)} = \frac{1}{2}

Hence, the correct answer is option: 'B'.

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