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BITSAT 2018
Exam
BITSAT
Year
2018
Subject
Physics
Chapter
Wave Optics

BITSAT 2018 · Physics · Wave Optics

In Young's double slit experiment, intensity at a point is...

In Young's double slit experiment, intensity at a point is (14)\left(\dfrac{1}{4}\right) of the maximum intensity. Angular position of this point is

Options

  1. Asin⁡−1(λd)\sin^{- 1} \left(\dfrac{λ}{d}\right)
  2. Bsin⁡−1(λ2d)\sin^{- 1} \left(\dfrac{λ}{2d}\right)
  3. Csin⁡−1(λ3d)\sin^{- 1} \left(\dfrac{λ}{3d}\right)
  4. Dsin⁡−1(λ4d)\sin^{- 1} \left(\dfrac{λ}{4d}\right)

Correct answer

C. sin⁡−1(λ3d)\sin^{- 1} \left(\dfrac{λ}{3d}\right)

Step-by-step solution

I = {I}_\max\cos^{2} \left(\dfrac{ϕ}{2}\right)

\dfrac{Imax}{4} = {I}_\max\cos^{2}\dfrac{ϕ}{2}

coscos\dfrac{ϕ}{2} = \dfrac{1}{2}

\dfrac{ϕ}{2} = \dfrac{\pi}{3}

⇒ϕ=⇒ ϕ =\dfrac{2\pi}{3} = \left(\dfrac{2\pi}{λ}\right) \times \Deltax . . . . \left( i\right)</p><p>where,</p><p>where,\Deltax = dsinθ</p><p>Puttingthedsin\theta</p><p>Putting thevalue of\Deltax inEq.(i),weget</p><p>sinθ=Eq. (i), we get</p><p>sin\theta =\dfrac{λ}{3d}</p><p>θ=sin⁡−1(λ3d)</p><p>\theta = \sin^{- 1} \left(\dfrac{λ}{3d}\right)

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