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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 bysin−1[μ1μ2cos(sin−1(μ2μ1))]sin−1[μ1cos(sin−1(1μ2))]sin−1(μ1μ2)sin−1(μ2μ1)

A
sin1[μ1μ2cos(sin1(μ2μ1))]
B
sin1(μ1μ2)
C
sin1[μ1cos(sin1(1μ2))]
D
sin1(μ2μ1)
Solution
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#### For the ray to emerge from the face CD, it at first must undergo total internal reflection at face AD i-e, r1>θcr1=sin−1(μ2μ1) .........................(1)Also from geometry, r=90−r1⟹sinr=cosr1 .............(2)Snell's law for face AB, μ2sinαmax=μ1 sinrThus from (1) & (2), μ2 sinαmax=μ1cos(sin−1(μ2μ1))Thus αmax=sin−1[μ1μ2cos(sin−1(μ2μ1))]

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