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As shown in figure a ray of light in air is incident at 30^@ on a medium and proceeds ahead in the medium. The refractive index of this medium varied with distance y as given by n(y) = 1.6 + (0.2)/(y +1)^(2) where y is in cm. What is the angle formed by the ray with the normal at a very large depth ? |
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Answer» Solution :`rArr` Suppose the angle is `theta` at distance in the MEDIUM. Appyling Snell.s law at this point, `N(y) SIN theta= C,` where C = constant....... (1) This formula is ture for all the pionts, Applying it to poin O, `n(0) sin 30^@ = C` But, `n(0) = 1.6 + (0.2)/((0+1))^2` `thereforen(0) = 1.8 ` From equation (2), `therefore n(0) = 1.8 xx 1/2 =C` `therefore 0.9 = C` Puting this value in (1), n(y) sin `theta` = 0.9 `therefore{1.6 + (0.2)/((y+1)^2)} sin theta =0.9` `therefore sin theta= (0.9)/(1.6 + (0.2)/(y+1)^2)` When y is very large, taking `y rarr oo`, we get sin `theta = (0.9)/(1.6)` `thereforetheta = 34^@ 14.` |
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