Question

One face of a prism of refracting angle 30° and refractive index 1.414 is silvered at what angle must a ray of light face on the unolivered face so that after refraction into the prism and reflection at the silver surface it retraces the path

Answer 1

Solution :

given A=30∘

μ=1.414

i2=0∘

because the refracted ray retracts its path

r2=0

r1+r2=A

r1+0=A

r1=30∘

μ=sini1sinr1

1.414=sini1sin30

sini1=1.414×12

=0.707

i=45∘

Answer 2

Answer :- 45°

Here A = 30° , i₂ = 0 since refracted the retraces the path, ∴ r₂ = 0 , μ = 1.414

now, r₁ + r₂ = 30°

Since r₂ = 0, then r₁ = 30°

μ = sin i₁ / sin r₁

∴ sin i₁ = 1.414 x sin 30° = 0.707

sin  i₁ = 0.707

Then i₁ = 45°