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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 . |
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Answer» Solution :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` 
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