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)
$x=\frac{(m+1)}{m}f+(m+1)f$
$f=\frac{mx}{(m+1{)}^2}$