JEE Main 2024 Physics question
A light ray is incident on a glass slab of thickness 4 cm and refractive index . The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is cm. (Given sin15º = 0.25)
Answer: 2
Solution
sin C = 1/√2 ⇒ i = 45°; sin r = sin 45°/√2 = 1/2 ⇒ r = 30°. Lateral shift = t sin(i − r)/cos r = 4√3 × 0.25/(√3/2) = 2 cm.