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NEET
Exam
NEET
Subject
Physics
Chapter
Wave Optics

NEET · Physics · Wave Optics

Given below are two statements: one is labelled as Assertion (A)...

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In Young's double slit experiment, the ratio of intensity at a point where path difference is λ4\dfrac{\lambda}{4} to the intensity at the central maximum is 1:21:2.

Reason (R): The phase difference corresponding to a path difference of λ4\dfrac{\lambda}{4} is π2\dfrac{\pi}{2}, and intensity varies as the square of the cosine of half the phase difference.

In the light of the above statements, choose the most appropriate answer from the options given below:

Options

  1. ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
  2. BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. C(A) is correct but (R) is not correct
  4. D(A) is not correct but (R) is correct

Correct answer

A. Both (A) and (R) are correct and (R) is the correct explanation of (A)

Step-by-step solution

Phase difference ϕ\phi is related to path difference Δx\Delta x by ϕ=2πλΔx\phi = \dfrac{2\pi}{\lambda} \Delta x.

For Δx=λ4\Delta x = \dfrac{\lambda}{4}, ϕ=2πλ×λ4=π2\phi = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{4} = \dfrac{\pi}{2}.

The intensity II at a point on the screen is given by I = I_\max \cos^2\left(\dfrac{\phi}{2}\right).

Substituting ϕ=π2\phi = \dfrac{\pi}{2}, we get I = I_\max \cos^2\left(\dfrac{\pi}{4}\right) = I_\max \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_\max}{2}.

The ratio of intensity at this point to the central maximum intensity is 1:21:2. Thus, Assertion (A) is correct.

Reason (R) correctly states the phase difference and the formula for intensity variation, which directly explains the calculation in the Assertion. Thus, Reason (R) is correct and is the correct explanation of (A).

Answer: Both (A) and (R) are correct and (R) is the correct explanation of (A)
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