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The angle of minimum deviation of a glas...

The angle of minimum deviation of a glass prism refraction angle `60^(@)` is `30^(@)` . If velocity of light in vacuum is `3 xx 10^(8) m * s^(-1)`, then determine its velocity glass.

Text Solution

Verified by Experts

Refraction index of material of the prism,
`mu = (sin""(A + delta_(m))/(2))/(sin(A)/(2))`
`therefore " " mu = ((sin""(60^(@) + 30^(@))/(2))/(2))/(sin""(60^(@))/(2)) = (sin45^(@))/(sin30^(@)) = ((1)/(sqrt(2)))/((1)/(2)) = sqrt(2)`
`"Again", mu = ("velocity of light in vacuum")/("velocity of light in glass"(v_(g)))`
`therefore " " sqrt(2) = (3 xx 10^(8))/(v_(g))`
`or, v_(g) = (3 xx 10^(8))/(sqrt(2)) = 2.12 xx 10^(8) m * s^(-1)`
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