Question

When a convex lens is placed in between the object and the screen, and the object is at a distance of $45\text{cm}$ from the screen by the convex lens, we obtain a small image of the object on the screen. By moving the lens, we receive a different image on the screen, whose size is $4$ times greater than the first. The focal length of the convex lens is ( in $\text{cm}$ )

• A. $9$
• B. $12$
• C. $10$
• D. $11$

Answer 1

The correct option is C $10$



As we know,

$h{{}_o}^2=h_{I_1}\times h_{I_2}......\left(1\right)$

Where,

$h_o-\text{Height of the object}{h}_{I_1}-\text{Height of the image in}{1}^{st}\text{case}{h}_{I_2}-\text{Height of the image in}{2}^{nd}\text{case.}$

Given,

$h_{I_2}=4h_{I_1}$

$\Rightarrow h{{}_o}^2=h_{I_1}\times 4h_{I_1}=4h_{I_1}^2$

$\Rightarrow h_o=2h_{I_1}$

$\therefore \frac{h_{I_1}}{h_o}=\frac{1}{2}......\left(2\right)$

From, the relation,

$m=\frac{h_{I_1}}{h_o}=\frac{v}{u}=\frac{45-u}{u}$

$\Rightarrow \frac{1}{2}=\frac{45-u}{u}$

$\Rightarrow 3u=90\Rightarrow u=30\text{cm}$

$v=D-u=45-30=15\text{cm}$

Now, Using focal length of the lens,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{(-u)}=\frac{1}{15}+\frac{1}{30}=\frac{2+1}{30}=\frac{3}{30}$

$f=10\text{cm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.