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A light beam is incident on a denser medium whose refractive index is 1.414 at an angle of incidence 45^@. Find the ratio of width of refracted beam in a medium to the width of the incident beam in air. |
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Answer» `sqrt3:sqrt2`  Widht of incident beam = 1 `triangle`ABC is RIGHT ANGLE triangle `therefore BC=SQRT(AB^2+AC^2)` `=sqrt(1+1)` `=sqrt2` Now, `angle`D is right angle in `triangle`BDC and at B and C `N=(sini)/(sinr)` `1.414=(sin45^@)/(sinr)` `therefore sin" "r=((1)/(sqrt2))/(sqrt2)=1/2` `therefore r=30^@` `therefore angleCBD=30^@` Now,in `triangle`BDC, `cos30^@=(BD)/(BC)=(x)/(sqrt2)` `(sqrt3)/(2)=(x)/(sqrt2)implies x=(sqrt3)/(sqrt2)` |
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