Q4. Assertion: in single-slit diffraction the first minimum occurs where the path difference between the edges of the slit is λ. Reason: the two halves of the slit then cancel point by point. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: with a sin θ = λ, each point of the upper half has a partner a/2 below it whose path differs by λ/2, so every pair cancels (reasoned from NCERT page 267).
Q5. The first minimum of a single-slit pattern is 5.0 mm from the centre on a screen 1.0 m away, for light of 500 nm. Find the slit width.
θ1 = 5.0 × 10−3/1.0 = 5.0 × 10−3 rad = λ/a, so a = (500 × 10−9)/(5.0 × 10−3) = 1.0 × 10−4 m = 0.10 mm (NCERT page 267).
Q6. A single-slit experiment is done in water (n = 4/3) instead of air. What happens to the width of the central maximum?
The wavelength in water is λ/n, so the width 2λD/a becomes 3/4 of its value in air (reasoned from NCERT Eq. (10.7), page 259; beyond the NCERT text).
Q7. For light of 600 nm through a slit 0.30 mm wide, at what angle is the second minimum, and roughly where is the first side maximum?
Second minimum: θ = 2λ/a = 2 × (600 × 10−9)/(0.30 × 10−3) = 4.0 × 10−3 rad. First side maximum: about 1.5λ/a = 3.0 × 10−3 rad (NCERT page 267).