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Question
Trace the path of a ray of light passing through a glass prism (ABC) as shown in the figure. If the refractive index of glass is `sqrt3`, find out the value of the angle of emergence from the prism.
Solution
Refractive index of glass `n_g =sqrt3`
Since i = 0
At the interface AC, we have according to shell’s law
`sin i/sinr = n_g/n_a`
But sin i = sin 0 = 0
Thus sin `r = (n_a sin i)/n_g =0`
Hence r = 0
This ray pass unrefracted at AC interface and reaches AB interface. Here we can see angle of incidence becomes 30°.
Thus, applying Snell’s law
`(sin30°)/(sin e) = n_a/n_g = 1/sqrt3`
`sin e = sqrt3 xx sin 30° = sqrt3/2`
Thus e = 60°
Hence, angle of emergence is 60°.
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