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Question

A ray of light is incident at the glass-water interface at an angle i = 41.8°. It emerges finally parallel to the surface of the water. What would be the value of the refractive index of the given glass? (Given: refractive index of water = 4/3)


(use the value: sin 41.8° ≈ 0.66)

[0.8 mark]

A
7/3
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B
1.5
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C
4/3
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D
1
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Solution

The correct option is B 1.5
Using Snell’s law at water-air interface,

sin rsin 90=1μwsin rsin 90=34

sin r = 3/4

Using Snell’s law at glass-water interface,
sin isin r=μwμg sin 41.83/4=(4/3)μg
μg=1.5

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