Wave Optics

JEE Main level, numerical answer, medium

33 of 191

Question 33 of 191

In a Young's double slit experiment, the intensity at some point on the screen is found to be 34\dfrac{3}{4} times of the maximum of the interference pattern. The path difference between the interfering waves at this point is λx\dfrac{\lambda}{x} where λ\lambda is wavelength of the incident light. The value of xx is _______.

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Answer: 6

The intensity at a point in Young's double slit experiment is given by I=Imaxcos⁡2(ϕ2)I = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)

Given I=34ImaxI = \dfrac{3}{4} I_{max}

34Imax=Imaxcos⁡2(ϕ2)\dfrac{3}{4} I_{max} = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)

cos⁡(ϕ2)=32\cos\left(\dfrac{\phi}{2}\right) = \dfrac{\sqrt{3}}{2}

ϕ2=π6\dfrac{\phi}{2} = \dfrac{\pi}{6}

ϕ=π3\phi = \dfrac{\pi}{3}

The phase difference ϕ\phi is related to the path difference Δx\Delta x by ϕ=2πλΔx\phi = \dfrac{2\pi}{\lambda} \Delta x

2πλΔx=π3\dfrac{2\pi}{\lambda} \Delta x = \dfrac{\pi}{3}

Δx=λ6\Delta x = \dfrac{\lambda}{6}

Comparing with Δx=λx\Delta x = \dfrac{\lambda}{x}, we get x=6x = 6

Answer: 66

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