Question

A convex lens forms an image of an object on a screen $30cm$ from the lens. When the lens is moved $90cm$ towards the object, then the image is again formed on the screen. Then, the focal length of the lens is

• A. $13cm$
• B. $24cm$
• C. $33cm$
• D. $40cm$

Answer 1

The correct option is B $24cm$

$\begin{array}{c}\text{From the Len's formulae, we know that}\\ \frac{1}{30}-\frac{1}{(-u)}=\frac{1}{4}\\ \frac{1}{30}+\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \Rightarrow \frac{1}{120}+\frac{1}{u-90}=\frac{1}{f}\\ \Rightarrow \frac{1}{30}+\frac{1}{u}=\frac{1}{120}+\frac{1}{u-a_0}\end{array}$

$\begin{array}{cc}\frac{1}{30}-\frac{1}{120}& =\frac{1}{u-a_0}-\frac{1}{u}\\ \frac{44}{120}& =\frac{4-\text{/}1+90}{u(u-90)}\\ u(u-a_0)& =40\times 90\\ & =120\times 30\\ \Rightarrow u=120cm& \end{array}$

$\text{Now}\begin{array}{c}\frac{1}{30}+\frac{1}{u}=\frac{1}{f}\\ \frac{1}{30}+\frac{1}{120}=\frac{1}{f}\\ f=\frac{120}{5}=24cm\end{array}$