Question

A convex lens of focal length f is placed some where in between an object and a screen. The distance between object and screen is x. If numerical value of magnification produced by lens is m, focal length of lens is

• A. $\frac{mx}{(m+1)^2}$
• B. $\frac{mx}{(m-1)^2}$
• C. $\frac{(m+1)^2}{m}x$
• D. $\frac{(m-1)^2}{m}x$

Answer 1

Answer: (A)

Here, x = u + v $ $...(i)
$ m =\frac{f}{f-u}=\frac{f-v}{f} $
Image is real, magnification is negative
$ - m =\frac{f}{f-u}$
$\Rightarrow u =\frac{-(m+1)}{m}f $
From, $ - m =\frac{f-v}{f}$
$\Rightarrow v = (m+1) f $
Put in Eq. (i)
$ x =\frac{(m+1)}{m}f+(m+1)f$
$ f =\frac{mx}{(m+1)^2}$