InterviewSolution
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InterviewSolution
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A ray of light travelling in air is incident at grazing angle (incident angle = 90^@) on a large rectangular slab of a transparent medium of thickness t = 1 m. The point of incidence js the origin A (0, 0). The medium has a refractive index of n (y) given n(y)= (9ky^(3/2)+ 1)^(1/2)Where K = ("meter")^((-3)/2), The refractive index of air is 1. Obtain the equation for the trajectory y(x) of the ray of the medium. |
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Answer» SOLUTION :If `theta` be the incident angle, the slope of the curve (TRAJECTORY) at any point, `m=tan(90^@-theta)=cot theta` Since, `(sintheta_(1))/(sintheta_(2)) = n_(2)/n_(1) implies n_(1) sintheta_(1)= n_(2)sintheta_(2)` If the RAY is incident from the air on the medium at an angle `theta_(1)=90^@` then `sintheta_(1)=1` and `n_(1)=1` `implies n_(2)sintheta_(2) =1 implies sintheta_(2)= 1/n` Putting `theta_(2)=theta` we obtain, `implies SIN theta=1/n implies cot theta = sqrt(n^(2)-1)` Since `m= dy/dx = cottheta implies dy/dx = sqrt(n^(2)-1)` Since `n= sqrt(ky^(3//2) +1)` we obtain `dy/dx = sqrt(ky^(3//2))` `4y^(1//4) = k^(1//2)x` 
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