Question

Focal length of the plane-convex lens is $15\text{cm}$. A small object is placed at $A$ as shown in the figure. The plane surface is silvered. The image will form at

• A. $12\text{cm}$ infront of the combination
• B. $24\text{cm}$ infront of the combination
• C. $10\text{cm}$ infront of the combination
• D. $18\text{cm}$ infront of the combination

Answer 1

The correct option is A $12\text{cm}$ infront of the combination

The equivalent focal length is given by,
$\frac{1}{F}=\frac{2}{f}+\frac{1}{f_m}$

$\frac{1}{F}=\frac{2}{15}+\frac{1}{\infty }$

$F=\frac{15}{2}\text{cm}$

Since $F$ has a positive value, the combination behaves as a converging mirror

Here, $u=-20\text{cm};F=\frac{-15}{2}\text{cm};v=?$

Using mirror formula, $\frac{1}{F}=\frac{1}{v}+\frac{1}{u}$

$\Rightarrow \frac{1}{(-\frac{15}{2})}=\frac{1}{v}-\frac{1}{20}$

$\Rightarrow \frac{1}{20}-\frac{2}{15}=\frac{1}{v}$

$\Rightarrow \frac{15-40}{20\times 15}=\frac{1}{v}$

$\Rightarrow v=\frac{20\times 15}{-25}$

$\Rightarrow v=-12\text{cm}$

The -ve sign indicates the image is infront of the mirror.