A monochromatic beam of light is incident at 60o on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal. For n=√3 the value of θ is 60o and dθdn=m. The value of m is
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Solution
From the geometry we have r1+r2=60o Using Snells law at the left side we have sin60osinr1=n or sinr1=√32n Using Snells law at the right end sinr2sinθ=1n or sinθ=nsinr2 =nxin(60o−r1) =n(√32cosr1−12sinr1) or =n(√32√1−sin2r1−12sinr1) or =n(√32√1−34n2−12√32n) Solving we get sinθ=√34√4n2−3−√34 Differentiating cosθdθ=√34×12√4n2−3×8ndn or dθdn=√3n√4n2−3×1cosθ =√3×√3√4×3−3×11/2 =2