A Young's double-slit experiment is performed using monochromatic light of wavelength λ. The intensity of light at a point on the screen, where the path difference is λ, is K units. The intensity of light at a point where the path difference is λ/6 is given by nK/12, where n is an integer. The value of n is __________.

Solution

Physics sir · 4 steps

Path difference λ means full brightness, so K is the maximum. Scale it by cos² of half the phase.

1. Intensity formula. For two equal slits, I = Iₘₐₓ cos²(φ/2), where the phase difference is φ = 2π/λ Δ x.
2. Find the maximum. At Δ x = λ, φ = 2π and cos² π = 1. So this point is a bright fringe and Iₘₐₓ = K.

   Δ x = λ/6 ⇒ φ = π/3 ⇒ I = K cos² π/6 = 3K/4

3. Match the form. Write 3K/4 with 12 in the denominator: 3K/4 = 9K/12.
4. So. Comparing with nK/12 gives n = 9.

The common slip: 3. Using cos² φ instead of cos²(φ/2) gives K cos² π/3 = K/4 = 3K/12, so n = 3, not 9.

The idea to keep. In YDSE, any path difference that is a whole number of wavelengths gives Iₘₐₓ. Elsewhere, use I = Iₘₐₓ cos²(φ/2), and halve the phase before taking the cosine.

Answer from the NTA final answer key. Solution by PrepWiser.