Wave Optics

JEE Main level, single correct, medium

30 of 191

Question 30 of 191

In a Young double slit experiment, the wavelength of incident light is 60006000 Å, the separation between slits S1S_1 and S2S_2 is 55 cm and the distance between slits plane and screen is 5050 cm, as shown in the figure below. If the resultant intensity at PP is equal to the intensity due to individual slits, the path difference between interfering waves is __________ Å.

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Answer: (C) 20002000

Let the intensity due to each individual slit be I0I_0.
The resultant intensity
IRI_R at a point where the phase difference is ϕ\phi is given by:

IR=I1+I2+2I1I2cos⁡ϕI_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi

Given that
I1=I2=I0I_1 = I_2 = I_0 and the resultant intensity IR=I0I_R = I_0, we can substitute these values into the equation:

I0=I0+I0+2I0⋅I0cos⁡ϕI_0 = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi

I0=2I0+2I0cos⁡ϕI_0 = 2I_0 + 2I_0 \cos \phi

I0=2I0(1+cos⁡ϕ)I_0 = 2I_0 (1 + \cos \phi)

1+cos⁡ϕ=121 + \cos \phi = \dfrac{1}{2}

cos⁡ϕ=−12\cos \phi = -\dfrac{1}{2}

The minimum phase difference satisfying this condition is:

ϕ=2π3\phi = \dfrac{2\pi}{3}

The relationship between phase difference
ϕ\phi and path difference Δx\Delta x is:
ϕ=2πλΔx\phi = \dfrac{2\pi}{\lambda} \Delta x

Substituting the value of
ϕ\phi:
2π3=2πλΔx\dfrac{2\pi}{3} = \dfrac{2\pi}{\lambda} \Delta x

Δx=λ3\Delta x = \dfrac{\lambda}{3}

Given the wavelength
λ=6000 A˚\lambda = 6000 \text{ \AA}:
Δx=60003=2000 A˚\Delta x = \dfrac{6000}{3} = 2000 \text{ \AA}

Answer:
20002000

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