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Question

An isosceles triangular glass prism stands with its base in water as shown. The angles that its two equal sides make with the base are θ each. An incident ray of light parallel to the water surface internally reflects at the glass-water interface and subsequently reemerges into the air. Take the refractive indices of glass and water to be 3/2 and 4/3 respectively. The least value of θ for this to happen must be


A
tan1217
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B
tan1817
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C
tan123
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D
tan183
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Solution

The correct option is A tan1217

Using Snell's law for initial refraction,

sin(90θ)=32sinr

cosθ=32sinr

sinr=23cosθ

For total internal reflection,

90x>c, where c is critical angle.

r+θ>c

(90x=r+θ from the geometry of the triangle)

sin(r+θ)>sinc

sinrcosθ+cosrsinθ>89

(μgw=32×34=89
sin c=1μgw=89)

23cos2θ+94cos2θ3sinθ>89

94cos2θsinθ>(832cos2θ)

(94cos2θ)(1cos2θ)>649+4cos4θ323cos2θ

cos2θ<1721sec2θ<2117tanθ>217

(tan2θ=sec2θ1)

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