Wave Optics

JEE Main level, numerical answer, medium

58 of 191

Question 58 of 191

In Young's double slit experiment, monochromatic light of wavelength 5000A∘5000A^{^{\circ}} is used. The slits are 1.0mm1.0mm apart and screen is placed at 1.0m1.0m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ______×10−6m.\times {10}^{-6}m.

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Answer: 125

Let intensity of light on screen due to each slit is I0{I}_{0}.
So internity at centre of screen is 4I04{I}_{0}(as cos⁡ϕ=1\cos \phi =1).
Intensity at distance yy from centre,
I=I0+I0+2I0I0cos⁡ϕI={I}_{0}+{I}_{0}+2\sqrt{{I}_{0}{I}_{0}}\cos \phi
As we know, Imax⁡=4I0{I}_{\max }=4{I}_{0}. Therefore,
⇒Imax⁡2=2I0=2I0+2I0cos⁡ϕ\Rightarrow \dfrac{{I}_{\max }}{2}=2{I}_{0}=2{I}_{0}+2{I}_{0}\cos \phi
⇒cos⁡ϕ=0\Rightarrow \cos \phi =0
⇒ϕ=π2\Rightarrow \phi =\dfrac{\pi }{2}
Hence, kΔx=π2k\Delta x=\dfrac{\pi }{2}
⇒2πλ(dsinθ)=π2\Rightarrow \dfrac{2\pi }{\lambda }\left(dsin\theta \right)=\dfrac{\pi }{2}
⇒2λd×yD=12\Rightarrow \dfrac{2}{\lambda }d\times \dfrac{y}{D}=\dfrac{1}{2}
⇒y=λD4d=5×10−7×14×10−3\Rightarrow y=\dfrac{\lambda D}{4d}=\dfrac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}
=125×10−6=125\times {10}^{-6}
=125=125

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