Question

Optical axis of a thin equi-convex lens is the $x-$ axis. The co-ordinates of a point object and its image are $(-40\text{cm},1\text{cm})$ and $(50\text{cm},-2\text{cm})$ respectively. Pole of the lens is located at :

[Assume that the lens is constrained to move only on $x-$ axis]

• A. $x+20\text{cm}$
• B. $x=-30\text{cm}$
• C. $x=-10\text{cm}$
• D. Origin

Answer 1

The correct option is C $x=-10\text{cm}$

Let us say the pole of the lens is located at $x\text{cm}$.

From the data given in the question,

Co-ordinates of object $(x_1,y_1)=(-40,1)\text{cm}$
Co-ordinates of image $(x_2,y_2)=(50,-2)\text{cm}$



From this,

Object distance $u=-(40+x)\text{cm}$
Image distance $v=(50-x)\text{cm}$

From the definition of lateral magnification,

$m=\frac{h_i}{h_o}=\frac{v}{u}$

Given,

$h_i=y_1=-2\text{cm}$

$h_o=y_2=1\text{cm}$

$\therefore \frac{-2}{1}=\frac{50-x}{-(40+x)}$

$\Rightarrow 80+2x=50-x$

$\Rightarrow 3x=-30\Rightarrow x=-10\text{cm}$

Hence, option (c) is the correct answer.