In a two-slit interference setup, if the slit separation d is halved while keeping λ and m constant, what happens to sin θ for the first bright fringe?

• A. It remains the same
• B. It doubles
• C. It halves
• D. It quadruples

Answer: (B)

Interference maxima occur when the path difference between the two slits equals an integer multiple of the wavelength: d sin θ = m λ. For the first bright fringe, m = 1, so sin θ = λ/d. If the slit separation is halved while keeping λ and m (and the order) fixed, sin θ becomes λ/(d/2) = 2λ/d, which is twice the original value. So sin θ doubles. (This assumes the angle remains physically possible, i.e., sin θ ≤ 1 for the setup.)