GATE 2018 IN Q.54: solution and answer

GATE 2018 · INQ.54 · 2 marks · NAT

A Michelson Interferometer using a laser source of wavelength $λ_{0} = 500$ nm, with both the mirrors ( $M_{1}$ & $M_{2}$ ) fixed and positioned equidistant from the splitter/combiner is shown in the figure. When a dielectric plate of refractive index $n = 1.5$ , of thickness $t$ , is placed in front of the mirror $M_{2}$ , a dark fringe is observed on the detector. When the dielectric plate is removed without changing the position of the mirrors $M_{1}$ & $M_{2}$ , a bright fringe is observed on the detector. The minimum thickness $t$ (in nm) of the dielectric is ________.

GATE 2018 IN question 54: Michelson interferometer. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

The minimum thickness of the dielectric plate is $t = 250 nm$ .

Official answer: 249 to 251Source: Official answer key, GATE 2018 IN (IIT Guwahati)

The mirrors $M_{1}$ and $M_{2}$ are equidistant from the splitter/combiner. With the plate removed, the two arms therefore have zero path difference, so the detector sees a bright fringe, as the question states.
With the plate in front of $M_{2}$ , the beam crosses it twice, once towards the mirror and once on the way back. In each pass the plate replaces a thickness $t$ of air by a thickness $n t$ of glass, so the optical path increases by $(n - 1) t$ per pass. The total extra optical path is $Δ = 2 (n - 1) t .$
The detector sees a dark fringe, which needs destructive interference. Since the arms were otherwise equal, the extra path must be an odd number of half-wavelengths: $2 (n - 1) t = (m + \frac{1}{2}) λ_{0}, m = 0, 1, 2, \dots$
The minimum thickness corresponds to $m = 0$ : $t_{m i n} = \frac{λ _{0}}{4 ( n - 1 )}$
Substitute $λ_{0} = 500 nm$ and $n = 1.5$ : $t_{m i n} = \frac{500 nm}{4 \times 0.5} = 250 nm$

KEY IDEA

A plate placed in one arm of a Michelson interferometer is crossed twice, so it adds an extra optical path of $2 (n - 1) t$ . Equal arms give a bright fringe, so a dark fringe needs this extra path to equal an odd number of half-wavelengths.

WATCH OUT FOR

Using a single pass, $(n - 1) t = λ_{0} /2$ , gives 500 nm. Another slip is using the full optical path $n t$ instead of the excess $(n - 1) t$ over air. Both ignore that the beam goes through the plate twice and that only the excess over air matters.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 11 Optical questions, with notes

GATE 2018 IN Q.54: solution and answer

Answer and solution

More Optical questions