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प्रश्न
Solve Numerical example.
A monochromatic ray of light is incident at 37° on an equilateral prism of refractive index 3/2. Determine angle of emergence and angle of deviation. If angle of prism is adjustable, what should its value be for emergent rays to be just possible for the same angle of incidence?
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उत्तर
By Snell’s law, in case of prism,
n = `sin("i")/sin("r"_1)`
∴ `3/2=sin(37°)/sin("r"_1)`
∴ r1 = `sin^-1(0.6018/(3//2))`
= sin−1 (0.4012)
= 23°39’
For equilateral prism, A = 60°
Also, A= r1 + r2
∴ r2 = A − r1 = 60° – 23°39’ = 36°21’
Applying snell’s law on the second surface of prism,
`1/"n"=sin("r"_2)/sin("e")`
∴ `2/3=sin(36°21’)/sin("e")`
∴ e = `sin^-1((0.5927)/(2//3))`
= sin–1 (0.889)
= 62°44’
≈ 63
For any prism,
i + e = A + δ
∴ δ = (i + e) − A
= (37 + 63) − 60
= 40°
For an emergent ray to just emerge, the angle r2’ acts as a critical angle.
∴ r2’ = `sin^-1(1/"n")`
= `sin^-1(2/3)`
= 41’48°
As, A = r1 + r2 and i to be kept the same.
⇒ A’ = r1 + r2’ = 23°39’ + 41°48’
= 65°27’
- Angle of emergence and angle of deviation in first case are 63° and 40° respectively.
- For emergent ray to just emerge with same incident angle the angle of prism should have the value of 65°27’.
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