Wave Optics

JEE Main level, single correct, medium

9 of 191

Question 9 of 191

In interference experiment the path difference between two interfering waves at a point AA on the screen is λ/3\lambda/3, where λ\lambda is the wavelength of these waves, and at another point BB the path difference is λ/6\lambda/6. The ratio of intensities at points AA and BB is _______.

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Answer: (C) 1/31/3

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

For point AA, the path difference is ΔxA=λ3\Delta x_A = \dfrac{\lambda}{3}.

ΔϕA=2πλ×λ3=2π3\Delta \phi_A = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3} = \dfrac{2\pi}{3}

The intensity at point AA is given by IA=Imaxcos⁡2(ΔϕA2)I_A = I_{max} \cos^2\left(\dfrac{\Delta \phi_A}{2}\right).

IA=Imaxcos⁡2(π3)=Imax(12)2=Imax4I_A = I_{max} \cos^2\left(\dfrac{\pi}{3}\right) = I_{max} \left(\dfrac{1}{2}\right)^2 = \dfrac{I_{max}}{4}

For point BB, the path difference is ΔxB=λ6\Delta x_B = \dfrac{\lambda}{6}.

ΔϕB=2πλ×λ6=π3\Delta \phi_B = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}

The intensity at point BB is given by IB=Imaxcos⁡2(ΔϕB2)I_B = I_{max} \cos^2\left(\dfrac{\Delta \phi_B}{2}\right).

IB=Imaxcos⁡2(π6)=Imax(32)2=3Imax4I_B = I_{max} \cos^2\left(\dfrac{\pi}{6}\right) = I_{max} \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_{max}}{4}

The ratio of intensities at points AA and BB is:

IAIB=Imax43Imax4=13\dfrac{I_A}{I_B} = \dfrac{\dfrac{I_{max}}{4}}{\dfrac{3I_{max}}{4}} = \dfrac{1}{3}

Answer: 1/31/3

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