GATE 2025 IN Q.55: solution and answer

GATE 2025 · INQ.55 · 2 marks · NAT

A schematic of a Michelson interferometer, used for the measurement of refractive index of gas, is shown in the figure. The transparent chamber is filled with a gas of refractive index $n_{g}$ , where $n_{g} \neq = 1$ , at atmospheric pressure. If a 532 nm laser beam produces 30 interference fringes on the screen, then the number of fringes produced by a 632.8 nm laser beam will be ______ (rounded off to one decimal place). Note: Assume that the effect of beamsplitter width is negligible. The setup is placed in air medium with refractive index equal to 1.

GATE 2025 IN question 55: Michelson interferometer. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

The 632.8 nm laser produces about 25.2 fringes.

Official answer: 24.5 to 25.5Source: Official answer key, GATE 2025 IN (IIT Roorkee)

The beam crosses the gas chamber twice (to the right-hand mirror and back), so the gas adds an optical path $Δ = 2 L (n_{g} - 1)$ compared with the same length of air, with $L = 2 cm$ . The two arms are otherwise identical in length and the beamsplitter width is neglected, so nothing else changes.
The fringe count is the extra path divided by the wavelength: $N = \frac{2 L ( n _{g} - 1 )}{λ}$ .
For $λ_{1} = 532 nm$ and $N_{1} = 30$ : $n_{g} - 1 = \frac{30 \times 532 \times 1 0 ^{- 9}}{2 \times 0.02} = 3.99 \times 1 0^{- 4}$ .
Take $n_{g}$ to be the same at 632.8 nm (dispersion neglected). The path change $2 L (n_{g} - 1)$ is then unchanged, so $N \propto \frac{1}{λ}$ .
$N_{2} = N_{1} \frac{λ _{1}}{λ _{2}} = 30 \times \frac{532}{632.8} \approx 25.22$ .
Rounded to one decimal place, $N_{2} \approx 25.2$ fringes.

KEY IDEA

The number of fringes shifted equals the extra optical path from the gas, $2 L (n_{g} - 1)$ , divided by the wavelength. With the gas and chamber unchanged, the count varies inversely with wavelength, so you do not need to find $n_{g}$ .

WATCH OUT FOR

Using a single pass, $L (n_{g} - 1)$ , instead of the double pass $2 L (n_{g} - 1)$ . The factor cancels in the ratio here, but it gives a wrong $n_{g}$ . Another slip is to invert the ratio and get $30 \times 632.8/532 \approx 35.7$ .

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 11 Optical questions, with notes

GATE 2025 IN Q.55: solution and answer

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