1.

Consider the figure of question 8, the anglephi for which minimum deviation is produced will be given by

Answer»

`COS^(2)phi=(MU^(2)+1)/3`
`cos^(2)phi=(mu^(2)-1)/3`
`sin^(2)phi=(mu^(2)+1)/3`
`sin^(2)phi=(mu^(2)-1)/3`

SOLUTION :B
`ddelta/dphi=-(4dalpha)/dphi +2`
`ddelta/dphi=0`
`impliesdalpha/dphi=1/2`
`alpha=sin^(-1)[1/mu sinphi]`
`cos^(2)phi=(mu^(2)-1)/3`


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