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QJEEAdv17Ph1887: JEE Advanced 2017 Physics - Class 12 JEEAdv17Ph1887 Solution: Two coherent monochromatic point sources and of wavelength are…

Two coherent monochromatic point sources S1{S_1} and S2{S_2} of wavelength λ=600 nm\lambda = 600\,nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d=1.8d=1.8 mm.mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Δθ.\Delta \theta . Which of the following options is/are correct? JEE Advanced 2017 Paper 2 Offline Physics - Wave Optics Question 16 English

Multiple Choice Options

  • A The angular separation between two consecutive bright spots decreases as we move from P1 to P2 along the first quadrant.
  • B At P2 the order of the fringe will be maximum.
  • C A dark spot will be formed at the point P2.
  • D The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000.

Step-by-Step Verified Solution

JEE Advanced 2017 Paper 2 Offline Physics - Wave Optics Question 16 English Explanation

At any point P on circumference, the path difference, Δ\Delta x = d sin θ\theta

At P1, θ\theta = 0 ∘^\circ ⇒\Rightarrow Δ\Delta x = 0

At P2, θ\theta = 90 ∘^\circ ⇒\Rightarrow Δ\Delta x = d

For constructive interference, Δ\Delta x = n λ\lambda

∴\therefore n=dλ=1.8 mm600 nm=1.8×10−3m600×10−9m=3000n = {d \over \lambda } = {{1.8\,mm} \over {600\,nm}} = {{1.8 \times {{10}^{ - 3}}m} \over {600 \times {{10}^{ - 9}}m}} = 3000

Since, n is integer, hence P2 corresponds to bright spot and also it corresponds to maximum order of fringe.

Now, Δ\Delta x = d sin θ\theta

d Δ\Delta x = d cos θ\theta d θ\theta or R λ\lambda = d cos θ\theta d θ\theta

∴\therefore dθ∝1cos⁡θd\theta \propto {1 \over {\cos \theta }}

Hence, as θ\theta increases, cos θ\theta decreases and consequently d θ\theta increases.

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