1.

A glass prism of angle 60^(@) and mu_(g) = 1.66 is immersed in a liquid of mu_(1) = 1.33. Find the angle of minimum deviation for a parallel beam of light passing through prism.

Answer»

`9.4^(@)`
`17.2^(@)`
`12.6^(@)`
`9.2^(@)`

Solution :(b) `.^(w)mu_(g)=(.^(a)mu_(g))/(.^(a)mu_(w))=(1.66)/(1.33)`
When prism is DIPPED in water and `delta_(m)`is MINIMUM DEVIATION, then
`.^(w)mu_(g)=(sin((A+deltam)/(2)))/((sinA)/(2))`
` = (1.66)/(1.33)`
Solving, we get, `delta_(m) = 17.2^(@)`


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