Q4. Assertion: at the Brewster angle the reflected and refracted rays are perpendicular. Reason: tan iB = μ. Are both true, and does the reason explain the assertion?
Both are true and they go together: with Snell’s law, sin iB = μ sin r, the condition tan iB = μ gives sin r = cos iB, so r = 90° − iB and the two rays are at right angles (reasoned).
Q5. Find the Brewster angle for diamond, refractive index 2.42 (NCERT Table 9.1, page 230).
iB = tan−1 2.42 ≈ 67.5° (Brewster’s law; beyond the NCERT text).
Q6. The Brewster angle for a medium is 57°. Find its refractive index and the speed of light in it.
μ = tan 57° ≈ 1.54, so v = c/μ ≈ (3.0 × 108)/1.54 ≈ 1.95 × 108 m/s (Brewster’s law; v = c/μ is NCERT Eq. (10.5), n2 = c/v2, page 259).
Q7. Reflected light from a glass window is viewed through a polaroid that is turned. What do you see if the light meets the glass at the Brewster angle, and near normal incidence?
At the Brewster angle the reflected light is fully polarised, so turning the polaroid takes it from bright to completely dark twice per turn (NCERT Summary, page 272). Near normal incidence the reflected light is only slightly polarised, so the brightness changes only a little (reasoned; beyond the NCERT text).
Q8. Light polarised in the plane of incidence falls on glass (μ = 1.5) at 56.3°. How much is reflected?
Essentially none in the ideal model. 56.3° is the rounded Brewster angle, so the in-plane light is all refracted (beyond the NCERT text).
Q9. Light reflected from glass at the Brewster angle has intensity I. What passes a polaroid whose axis is at 60° to its field?
Malus’ law, NCERT Eq. (10.18) on page 270, gives I cos2 60° = I/4.
Q10. The critical angle from a material into air is 45°. Find the Brewster angle for incidence from air into that material.
tan iB = 1/sin 45° = √2, so iB ≈ 54.7° (reasoned).