Q132 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
The width of the fringes obtained with a light of wave length $6.2 \times 10^{-8}$ m is 1.82 mm. If the whole apparatus is immersed in a liquid of refractive index 1.3, what will be the width of the resulting fringe?
- Correct answerA$1.4$ mm
- B$1.82$ mm
- C$2.8$ mm
- D$0.71$ mm
Solution
The fringe width in air is given by $\beta = \dfrac{\lambda D}{d}$.
When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light in the liquid becomes $\lambda' = \dfrac{\lambda}{\mu}$.
The new fringe width is $\beta' = \dfrac{\lambda' D}{d} = \dfrac{\lambda D}{\mu d} = \dfrac{\beta}{\mu}$.
Given $\beta = 1.82$ mm and $\mu = 1.3$.
Substituting the values, $\beta' = \dfrac{1.82}{1.3} = 1.4$ mm.
Answer: $1.4$ mm
When the apparatus is immersed in a liquid of refractive index $\mu$, the wavelength of light in the liquid becomes $\lambda' = \dfrac{\lambda}{\mu}$.
The new fringe width is $\beta' = \dfrac{\lambda' D}{d} = \dfrac{\lambda D}{\mu d} = \dfrac{\beta}{\mu}$.
Given $\beta = 1.82$ mm and $\mu = 1.3$.
Substituting the values, $\beta' = \dfrac{1.82}{1.3} = 1.4$ mm.
Answer: $1.4$ mm
Hence, the correct answer is option: 'A'.