Wave Optics

JEE Main level, single correct, medium

31 of 191

Question 31 of 191

The maximum intensity in a Young's double slit experiment is I0I_0. Distance between the slits (dd) is 5λ5\lambda, where λ\lambda is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10dD = 10d is _______.

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Answer: (B) I02\dfrac{I_0}{2}

The position of the point on the screen exactly opposite to one of the slits is at a distance y=d2y = \dfrac{d}{2} from the central maximum.

The path difference Δx\Delta x at this point is given by:
Δx=ydD\Delta x = \dfrac{yd}{D}

Substituting y=d2y = \dfrac{d}{2} and D=10dD = 10d:
Δx=(d2)d10d=d220d=d20\Delta x = \dfrac{\left(\dfrac{d}{2}\right)d}{10d} = \dfrac{d^2}{20d} = \dfrac{d}{20}

Given that the distance between the slits is d=5λd = 5\lambda, we have:
Δx=5λ20=λ4\Delta x = \dfrac{5\lambda}{20} = \dfrac{\lambda}{4}

The corresponding phase difference Δϕ\Delta \phi is:
Δϕ=2πλΔx=2πλ(λ4)=π2\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi}{\lambda} \left(\dfrac{\lambda}{4}\right) = \dfrac{\pi}{2}

The intensity at this point is given by:
I=I0cos⁡2(Δϕ2)I = I_0 \cos^2\left(\dfrac{\Delta \phi}{2}\right)

Substituting Δϕ=π2\Delta \phi = \dfrac{\pi}{2}:
I=I0cos⁡2(π4)=I0(12)2=I02I = I_0 \cos^2\left(\dfrac{\pi}{4}\right) = I_0 \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{2}

Answer: I02\dfrac{I_0}{2}

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