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Exam
NEET
Subject
Physics
Chapter
Wave Optics

NEET · Physics · Wave Optics

In a Young's double slit experiment, the separation between the two...

In a Young's double slit experiment, the separation between the two slits is d=λ3d = \dfrac{\lambda}{3}, where λ\lambda is the wavelength of the monochromatic light used. If the intensity of the central maximum is I0I_0, the intensity of light at an angular position of 30∘30^\circ on the screen will be :

Options

  1. AI04\dfrac{I_0}{4}
  2. BI02\dfrac{I_0}{2}
  3. C3I04\dfrac{3I_0}{4}
  4. DZero

Correct answer

C. 3I04\dfrac{3I_0}{4}

Step-by-step solution

The path difference Δx\Delta x at an angular position θ\theta is given by Δx=dsin⁡θ\Delta x = d \sin\theta.

Given d=λ3d = \dfrac{\lambda}{3} and θ=30∘\theta = 30^\circ:

Δx=(λ3)sin⁡(30∘)=λ3×12=λ6\Delta x = \left(\dfrac{\lambda}{3}\right) \sin(30^\circ) = \dfrac{\lambda}{3} \times \dfrac{1}{2} = \dfrac{\lambda}{6}

The corresponding phase difference ϕ\phi is:

ϕ=2πλΔx=2πλ×λ6=π3\phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}

The intensity II at this point is given by I=I0cos⁡2(ϕ2)I = I_0 \cos^2\left(\dfrac{\phi}{2}\right), where I0I_0 is the intensity of the central maximum.

Substituting the value of ϕ\phi:

I=I0cos⁡2(π6)=I0(32)2=3I04I = I_0 \cos^2\left(\dfrac{\pi}{6}\right) = I_0 \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_0}{4}

Answer: 3I04\dfrac{3I_0}{4}
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