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प्रश्न
A ray of light passes through a prism of refractive index `sqrt2` as shown in the figure. Find:
- The angle of incidence (∠r2) at face AC.
- The angle of minimum deviation for this prism.
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उत्तर
(i) Since at point N, the angle of refraction is 90°, then ∠r2 is the critical angle for the glass-air pair of media.
sin ∠r2 = `1/μ = 1/sqrt2`
∴ ∠r2 = `sin^-1(1/sqrt2)` = 45°
(ii) μ = `sin(("A" + δ_"m")/2)/(sin "A"/2)`
Or, `sqrt2 = sin ((60^circ + δ_"m")/2)/(sin 60^circ/2)`
Or, `sqrt2 = sin (30^circ + δ_"m"/2)/(1/2)`
Or, 0.7 = `sin(30^circ + δ_"m"/2)`
Or, sin−1 0.7 = `30^circ + δ_"m"/2`
Or, 44.4° = `30^circ + δ_"m"/2`
∴ δm = 28.8°
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