Wave Optics

JEE Main level, single correct, medium

106 of 191

Question 106 of 191

In a Young's double slit experiment, slits are separated by 0.5mm0.5mm, and the screen is placed 150cm150cm away. A beam of light consisting of two wavelengths, 650nm650nm and 520nm520nm, 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:

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Answer: (C) 7.8mm7.8mm

Given,
Wavelength of first light is, λ1=650nm{\lambda }_{1}=650nm
Wavelength of second light is, λ2=520nm{\lambda }_{2}=520nm
Screen distance is, D=150cmD=150cm
Slit width is, d=0.5mmd=0.5mm

For first wavelength, location of fringe from central line,
y1=n1λ1Dd{y}_{1}=\dfrac{{n}_{1}{\lambda }_{1}D}{d} for bright fringes

For second wavelength, location of fringe from central line,
y2=n2λ2Dd{y}_{2}=\dfrac{{n}_{2}{\lambda }_{2}D}{d} for bright fringes

Condition of coincide of fringe is,
y1=y2{y}_{1}={y}_{2}
⇒n1λ1=n2λ2\Rightarrow {n}_{1}{\lambda }_{1}={n}_{2}{\lambda }_{2}
⇒n1×650=n2×520\Rightarrow {n}_{1}\times 650={n}_{2}\times 520
⇒n1n2=45\Rightarrow \dfrac{{n}_{1}}{{n}_{2}}=\dfrac{4}{5}

For the minimum value of n1{n}_{1} and n2{n}_{2}
n1=4{n}_{1}=4 and n2=5{n}_{2}=5
(y1)min=4×650nm×1.50m0.5×10−3m{\left({y}_{1}\right)}_{min}=4\times 650nm\times \dfrac{1.50m}{0.5\times {10}^{-3}m}
=7800×10−6m=7800\times {10}^{-6}m
=7.8mm=7.8mm

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