Initially, an object is kept at a distance of $10cm$ from the convex lens and a sharp image is formed at $10cm$ ahead of the lens on the screen. Now a glass plate of$\mu =1.5cm$ and thickness$1.5cm$ is placed between object and lens. The distance by which the screen be shifted to get a sharp image on the screen will be:

The correct option is B

$0.55cm$ away from the lens

Solution:

Step 1: Given information

Object distance, $u=10cm$

Image distance, $v=10cm$

The thickness of glass, $d=1.5cm$

Refractive index, $\mu =1.5$

Step 2: Calculate the distance by which the screen be shifted to get a sharp image on the screen

$$\begin{gathered} \Rightarrow 2f=10 \\ \Rightarrow f=5cm \end{gathered}$$

Therefore,

$$\begin{gathered} S=t\left[1–\left(\frac{1}{u}\right)\right] \\ S=1.5\left[1–\left(\frac{1}{1.5}\right)\right] \\ S=0.5 \end{gathered}$$

Step 3: Solving with the help of the lens formula

Along incident rays is positive,

New object distance$=10-0.5=9.5$

$\left(\frac{1}{v}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{f}\right)$

$$\begin{gathered} \Rightarrow V=\frac{uf}{(u+f)} \\ \Rightarrow V=\frac{(5x9.5)}{(9.5–5)} \\ \Rightarrow V=10.55cm \end{gathered}$$

The screen should be shifted by a distance$0.55cm$ away from the lens.

Hence, the correct option is (B).