Two slits separated by d = 0.25 mm produce interference with light λ = 500 nm. What is the approximate angle for the first maximum (m = 1)?

• A. About 0.23°
• B. About 0.11°
• C. About 1.1°
• D. About 0.04°

Answer: (B)

When bright fringes appear in a double-slit setup, the path difference between the two slits must be an integer multiple of the wavelength: d sin θ = m λ. For small angles, sin θ ≈ θ (in radians), so the angle of the m-th maximum is θ ≈ m λ / d.

Plugging in the numbers for the first maximum (m = 1): d = 0.25 mm = 2.5 × 10^-4 m and λ = 500 nm = 5 × 10^-7 m. So θ ≈ (1)(5 × 10^-7 m) / (2.5 × 10^-4 m) = 2 × 10^-3 radians. Converting to degrees, θ ≈ 2 × 10^-3 × (180/π) ≈ 0.11°. This matches the provided answer.

The small-angle approximation is valid here because the angle is only about a tenth of a degree. If you check other angles, they don’t satisfy d sin θ = λ for the first maximum: too large or too small angles would correspond to higher-order maxima or no constructive interference at m = 1.