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Question
Figures (a) and (b) show the refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in the glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Figure (c)].
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| (a) | (b) | (c) |
Solution
As per the given figure, for the glass-air interface:
Angle of incidence, i = 60°
Angle of refraction, r = 35°
The relative refractive index of glass with respect to air is given by Snell’s law as:
`""^"a"μ_"g" = (sin "i")/(sin "r")`
= `(sin 60°)/(sin 35°)`
= `(0.8660)/(0.5736)`
= 1.51 ......(1)
As per the given figure, for the air-water interface:
Angle of incidence, i = 60°
Angle of refraction, r = 47°
The relative refractive index of water with respect to air is given by Snell’s law as:
`""^"a"μ_"w" = (sin "i")/(sin "r")`
= `(sin 60°)/(sin 47°)`
= `(0.8660)/(0.7314)`
= 1.184 ....(2)
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
`""^"w"μ_"g" = (""^"a"μ_"g")/(""^"a"μ_"w")`
= `1.51/1.184`
= 1.275
The following figure shows the situation involving the glass-water interface.
Angle of incidence, i = 45°
Angle of refraction = r
From Snell’s law, r can be calculated as:
`(sin "i")/(sin "r") = ""^"w"μ_"g"`
`(sin 45°)/sin "r"` = 1.275
sin r = `(1/(sqrt2))/1.275`
sin r = 0.5546
∴ r = `sin^(-1) (0.5546)` = 33.68°
Hence, the angle of refraction at the water-glass interface is 33.68°.
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