ABCD is the plane of glass cube. A horizontal beam of light enters the face AB at the grazing incidence. Show that the angle θ which any ray emerging from BC would make with normal to BC is given by ⇒ sin θ = cot α
where α is the critical angle. What is the greatest value that the refraction index of glass may have if any of the light is to emerge from BC?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. At interface AB:
1 × sin 
= µ g . sin α
or sin α = 
… (1)
At interface BC:
µ g . sin (90 – α ) = 1 × sin θ
or sin θ = µ g cos α … (2)
From equation (1) and (2)
sin θ = cot α
For emergence from BC
90 – α ≤ C
Now at grazing incidence α = C
∴ 90 – α ≤ C
or 2 C ≥ 90º
Hence, greatest value of R.I. is 
or C ≥ 45º
or sin 
≥ 45º or 
≥ 
.
μ ≤ 
.
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