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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2026 · 28 January · Evening Shift · Physics · Wave Optics

A beam of light consisting of wavelengths 650 nm and 550 nm...

A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ____\_\_\_\_ ×10−5 m\times 10^{-5} \mathrm{~m}.

Correct answer

429

Step-by-step solution

Given wavelengths are λ1=650\lambda_1 = 650 nm and λ2=550\lambda_2 = 550 nm.
Slit separation d=2d = 2 mm =2×10−3= 2 \times 10^{-3} m.
Distance to screen D=1.2D = 1.2 m.
For the bright fringes of both wavelengths to coincide at a distance yy from the central maximum, we have:
y=n1λ1Dd=n2λ2Ddy = n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d}
where n1n_1 and n2n_2 are integers representing the order of the fringes.
This simplifies to n1λ1=n2λ2n_1 \lambda_1 = n_2 \lambda_2.
n1n2=λ2λ1=550650=1113\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}
The least values for coincidence are n1=11n_1 = 11 and n2=13n_2 = 13.
The least distance yy is given by:
y=n1λ1Dd=11×650×10−9×1.22×10−3y = n_1 \frac{\lambda_1 D}{d} = 11 \times \frac{650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}
y=11×650×0.6×10−6y = 11 \times 650 \times 0.6 \times 10^{-6} m
y=11×390×10−6y = 11 \times 390 \times 10^{-6} m
y=4290×10−6y = 4290 \times 10^{-6} m
y=429×10−5y = 429 \times 10^{-5} m.
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