Question

A prism has refracting angle equal to `pi//2`. It is given that `gamma` is the angle of minimum deviation and `beta` is the deviation of the ray entering at grazing incidence. Prove that `singamma=sin^(2)beta` .

Answer 1

Applying condition of minimum deviation,
`mu=" sin "(((A+gamma))/2)/("sin"A/2)=("sin"(A)/(2)"cos"(gamma)/(2)+"cos"(A)/(2)"sin"(gamma)/(2))/("sin"(A)/(2))`
`="cos" (gamma)/(2)+"cot" (A)/(2)"sin"(gamma)/(2)`
Using `A=90^(@), mu=" cos "(gamma)/(2)+cot45^@" sin "(gamma)/(2)`
`rArr " cos " (gamma)/(2)+ " sin "(gamma)/(2)=mu`
Squaring, `"cos"^(2)(gamma)/(2)+"sin"^(2)(gamma)/(2)+gamma=mu^(2)rArrsingamma=mu^(2)-1`
Deviation at grazing incidence,
`beta= delta_(1)+delta_(2)`
`beta=((pi)/(2)-C)+(e-r_(2))`
`rArr beta ((pi)/(2)-C)+[e-((pi)/(2)-C)]`
`rArr beta=e`
(iii) or `sin beta=sin e=mu sin r_(2)=mu sin((pi)/(2)-C)`
`rArr sin beta=mu cosC`
Squaring Eq. (ii),
`sin^(2) beta=mu^(2)cos^(2)C rArr sin^(2) beta=mu^(2)(1-sin^(2)C)`
Using `sin C = (1)/(mu), sin^(2)beta=mu^(2)(1-(1)/(mu^(2)))`
`rArr sin^(2)beta=mu^(2)-1`
From Eqs. (i) and (iii),
`sin gamma=sin^(2) beta`