Question

The speed of light in glass is $2\times {10}^8\frac{m}{s}$ and in the air is $3\times {10}^8\frac{m}{s}$ . An inkdot is covered by a glass plate of $6cm$ thickness. Calculate the shift in the ink dot where it appears to be raised from the original position.

Answer 1

Step-1: The given data -

Speed of light in the glass, v = $2\times {10}^8\frac{m}{s}$

Speed of light in air, c = $3\times {10}^8\frac{m}{s}$

The thickness of glass, t = $6cm$

Shift in the inkdot, s = ?

Step-2: Calculating the shift -

Refractive index, $n=\frac{c}{v}=\frac{3\times {10}^8}{2\times {10}^8}=1.5$

Also, the Refractive index, $$\begin{gathered} n=\frac{Realdepth\left(t\right)}{Apparentdepth\left(d\right)} \\ \to 1.5=\frac{6}{d} \\ \to d=4cm \end{gathered}$$

So, the shift in the inkpot, s = Real depth - Apparent depth = $6-4=2cm$

Hence, the shift in the inkpot is $2cm$.