Question

The graph shows the variation of magnification $m$ produced by convex lens with image distance $\vartheta$. The focal length of the lens used is :

• A. $\frac{b}{c}$
• B. $\frac{b}{ca}$
• C. $\frac{bc}{a}$
• D. $\frac{c}{b}$

Answer 1

The correct option is D $\frac{c}{b}$

$m=b=\frac{-\vartheta }{u}$

$\frac{1}{u}=\frac{m}{\vartheta }=\frac{b}{c}$

$u=\frac{b}{(a+c)}$

$\frac{1}{u}=\frac{m}{\vartheta }$

$\frac{1}{(a+c)}-\frac{b}{c}=\frac{1}{f}$

$\frac{1}{u}=\frac{b}{c}=\frac{b}{a+c}$

$1\Rightarrow \frac{1}{f}=\frac{b}{c}$

$\frac{b}{c}=\frac{b}{a+c}$

$\frac{1}{a+c}=\frac{2b}{c}$

$f=\frac{c}{b}$