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A rectangular glass slab ABCD of refractive index μ1 is immersed in water of refractive index μ2(μ2<μ1). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax such that the ray comes out from the other surface CD is given by
161972_368abb9dcf474a38826f2317ea601d92.png
  1. sin1[μ1μ2cos(sin1(μ2μ1))]
  2. sin1[μ1cos(sin1(1μ2))]
  3. sin1(μ1μ2)
  4. sin1(μ2μ1)

A
sin1[μ1μ2cos(sin1(μ2μ1))]
B
sin1(μ1μ2)
C
sin1[μ1cos(sin1(1μ2))]
D
sin1(μ2μ1)
Solution
Verified by Toppr

For the ray to emerge from the face CD, it at first must undergo total internal reflection at face AD i-e, r1>θc
r1=sin1(μ2μ1) .........................(1)
Also from geometry, r=90r1sinr=cosr1 .............(2)
Snell's law for face AB, μ2sinαmax=μ1 sinr
Thus from (1) & (2), μ2 sinαmax=μ1cos(sin1(μ2μ1))
Thus αmax=sin1[μ1μ2cos(sin1(μ2μ1))]

412765_161972_ans_5c550621210f4e7d95045d29f63e04a9.png

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