Question

A convex lens forms a real image three times larger than the object on a screen. Object and screen are moved until the image formed becomes twice the size of the object. If the shift of the object is $6\text{cm}$ then find the distance by which screen is shifted.

• A. $36\text{cm}$
• B. $72\text{cm}$
• C. $18\text{cm}$
• D. $9\text{cm}$

Answer 1

The correct option is A $36\text{cm}$

From the data given in the question, we can draw a schematic diagram of image formation as given below.


Using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

With sign convention, let us substitute the data given in the diagram,

$\frac{1}{3x}-\frac{1}{-x}=\frac{1}{f}$

or, $\frac{4}{3x}=\frac{1}{f}.....\left(1\right)$

Similarly,

$\frac{1}{2(6+x)}-\frac{1}{-(6+x)}=\frac{1}{f}$

or , $\frac{3}{2(6+x)}=\frac{1}{f}....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{4}{3x}=\frac{3}{2(6+x)}$

$\Rightarrow 9x=48+8x$

$\Rightarrow x=48\text{cm}$

Shift of screen $\left(x^{\prime }\right)=|2(6+x)-3x|$

Substituting the value of $x$ we get,

$x^{\prime }=3\times 48-2(6+48)=36\text{cm}$

Hence, option (a) is the correct answer.