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Question: Answered & Verified by Expert
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:
PhysicsWave OpticsJEE MainJEE Main 2017 (02 Apr)
Options:
  • A 15.6 mm
  • B 1.56 mm
  • C 7.8 mm
  • D 9.75 mm
Solution:
1142 Upvotes Verified Answer
The correct answer is: 7.8 mm

Given, 
Wavelength of first light is, λ1=650 nm
Wavelength of second light is, λ2=520 nm
Screen distance is, D=150 cm
Slit width is, d=0.5 mm

For first wavelength, location of fringe from central line,
y1=n1λ1Dd for bright fringes

For second wavelength, location of fringe from central line,
y2=n2λ2Dd for bright fringes

Condition of coincide of fringe is, 
y1=y2
n1λ1=n2λ2
n1×650=n2×520
n1n2=45

For the minimum value of n1 and n2
n1=4 and n2=5
y1min=4×650 nm×1.50 m0.5×10-3 m
=7800×10-6 m
=7.8 mm

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