JEE Main 2017 — Wave Optics Question with Solution
From: JEE Main 2017 (Offline)
Question
In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes
on the screen. The least distance from the common central maximum to the point where the bright fringes
due to both the wavelengths coincide is
Choose an option
Show full solutionCorrect option: C
Correct answer
C7.8 mm
Step-by-step explanation
Let n1th fringe formed due to first wavelength and n2th fringe formed due to second wavelength coincide. So their distance from common central maxima will be same.
yn1 = yn2
=
Hence, distance of the point of coincidence from the central maxima is
y = = = 7.8 mm
yn1 = yn2
=
Hence, distance of the point of coincidence from the central maxima is
y = = = 7.8 mm
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