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

The correct option is A $\frac{mx}{(m+1{)}^2}$


x = u + v …(i)
$m=\frac{f}{f-u}\Rightarrow \frac{f-v}{f}$
Image is real, magnification is negative
$-m=\frac{f}{f-u}u=\left(\frac{m+1}{m}\right)f-m=\frac{f-v}{f}x=\left[\frac{m+1}{m}\right]f+(m+1)f\Rightarrow f(m+1)[\frac{1}{m}+1]f=\frac{mx}{(m+1{)}^2}$