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Critical angle of glass and air is 42^(@) . If a ray of light is incident normally on a face of an equilateral prism, show that the ray will emerge from the base of the prism normally. Calculate the deviation of the ray. |
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Answer» Solution :ABC is an EQUILATERAL prism [FIG. 2.72]. Each ANGLE of the prism is `60^(@)`. A ray of light PQ is incident normally on the face AB.Without changing the direction it travels along QR and is incident at R on the face AC. According to the figure, `angleARQ = 30^(@)` ` .therefore" " "Angle of incidence at"R= 60^(@)` We know that critical angle of glass and air is `42^(@)`. So the ray incident at R will be totally reflected and will follow the path RST. `therefore " " angleNRS = 60^(@) "s", angleCRS = 30^(@)` `"Again" " " angleACB = 60^(@) "so," angleCSR = 90^(@)` So the ray RST will emerge along the normal to the face BC. The angle of DEVIATION of the ray = `180^(@) - (60^(@) + 60^(@)) = 60^(@)`. 
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