Q109 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
A glass slab of thickness $6.0$ cm is placed on the piece of paper on which an inkdot is marked. By how much distance would an inkdot appear to be raised? The velocity of light in glass is $2 \times 10^8 \text{ ms}^{-1}$ and that in air is $3 \times 10^8 \text{ ms}^{-1}$.
- Correct answerA$2.0$ cm
- B$3.0$ cm
- C$4.0$ cm
- D$5.0$ cm
Solution
The refractive index of glass is given by $\mu = \dfrac{c}{v_g}$
$\mu = \dfrac{3 \times 10^8}{2 \times 10^8} = 1.5 = \dfrac{3}{2}$
The apparent shift of the inkdot is given by the formula $\Delta x = t \left(1 - \dfrac{1}{\mu}\right)$
Substituting the given values:
$\Delta x = 6.0 \left(1 - \dfrac{2}{3}\right)$
$\Delta x = 6.0 \times \dfrac{1}{3} = 2.0 \text{ cm}$
The inkdot appears to be raised by $2.0 \text{ cm}$.
Answer: $2.0$ cm
$\mu = \dfrac{3 \times 10^8}{2 \times 10^8} = 1.5 = \dfrac{3}{2}$
The apparent shift of the inkdot is given by the formula $\Delta x = t \left(1 - \dfrac{1}{\mu}\right)$
Substituting the given values:
$\Delta x = 6.0 \left(1 - \dfrac{2}{3}\right)$
$\Delta x = 6.0 \times \dfrac{1}{3} = 2.0 \text{ cm}$
The inkdot appears to be raised by $2.0 \text{ cm}$.
Answer: $2.0$ cm
Hence, the correct answer is option: 'A'.