Question

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

• A. $60\text{cm}\text{to the left of lens}$
• B. $12\text{cm}\text{to the left of lens}$
• C. $60\text{cm}\text{to the right of lens}$
• D. $30\text{cm}\text{to the right of lens}$

Answer 1

The correct option is B $12\text{cm}\text{to the left of lens}$

Given, $f=15\text{cm};u=-20\text{cm}$



(i.) Refraction through the lens :

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

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

$\Rightarrow v=60\text{cm}$

(ii.) Reflection by the plane mirror :

After, reflection from the plane mirror, itsimage will be formed at $v_1=60\text{cm}$ to the left of the system. Which acts as a virtual object for the refraction through lens.

(iii.) Refraction through the lens :

$\frac{1}{f}=\frac{1}{v_1}-\frac{1}{u_1}$

$\frac{1}{15}=\frac{1}{v_1}-\frac{1}{60}$

$\Rightarrow v_1=12\text{cm}$

$\therefore$ The final image is formed at $12\text{cm}$ left to the system.

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