InterviewSolution
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InterviewSolution
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The mixture a pure liquid and a solution in long vertical column (i.e, horizontal dimensions lt lt vertical dimensions ) Produce diffusion of solute particles and hence refractive index gradient along the vertical dimesion . A ray of light entering the column I at right angles to the vertical is deviated from its original path. Find the deviated from its original path. Find the deviation in travelling a horizontal distance d lt lt h, the height of the coloumn. |
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Answer» Solution :`rArr` As shown in the figure, consider an extremely narrow region of width dx, between the layers at distances x and x + dx , inside the extremely high cylindrical columnof liquid . `rArr` In above region, point B is on the level `bar(PQ)` , at height y from horizontal reference level, where REFRACTIVE index is `mu` and gradient of refractiveindex. is `(dmu)/(dy)`. At this point, light ray `vec (AB)` is incident at angle `(180^@ - theta)` `therefore vec (AB)` , makes angle `(180^@ - theta)`with normal`M_1N_1` drawn on horizontal surface `bar(PQ)` , which becomes angle of incidence. ) If there would not be any gradient of refractive index, ray `vec(AB)` would have come to point B.. But here refractive index increases with the decreases in height and so at level `bar(RS)` height is (y - dy) andrefractive indes is `(mu + dmu)` which is more than `mu`. Hence, ray `vec(AB)`bends towards normal `M_1 N_1` and so it advances from B to C. Thus, ray `vec(BC)` becomes refracted light ray whichmakes angle `{180^@ - (theta + d theta)}` with normal `M_1 N_1`. Now applying Snell.s law atpoint B ion above figure, we get, `mu XX sin (180^@ - theta) = (mu + d mu) xx sin { 180^@ - {theta + d theta)}` `because mu sin theta = (mu + d mu) sin (theta+ d theta)` `thereforemu sin theta = (mu + d mu )(sin theta cos d theta + cos theta sin d theta)` ` therefore mu sin theta = mu sin theta cos (d theta) + mu cos theta sin (d theta) + d(mu ) sin theta cos (d theta) + (d mu) cos theta sin (d theta)` `rArr` Since `d theta` is extremely small, we can take sin `(d theta)` `= d ( theta) cos (d theta) =1` `mu sin theta = (mu sin theta xx 1) + ( mu cos theta xx d theta)` `therefore0 = mu cos theta d theta + (d mu) sin theta ( because d theta cos theta d mu ` are negligible) `therefore ( d mu) sin theta = - mu cos theta d theta` `therefore (d mu ) sin theta = - mu cos theta d theta` `therefore ( dmu ) tan theta = -mu d theta`.........(1) `rArr` For the right angled `DELTA` BEB. in the figure `tan(180^@-B) = (BE)/(EB.)` `therefore tan theta= (dx)/(dy)`..........(2) `rArr` From equation (1) and (2,) `(d mu)(dx/dy) = - mu d theta` `therefore d theta = - (1)/(mu) (d mu)/(dy) dx` `rArr` Taking INTEGRATION on both the sides, `int_(0)^(@) = - (1)/(mu) ((dmu)/(dy)) int_(0)^(d) dx` `(therefore`During horizontal displacement `mu " and " (dmu)/(dy)` remain constant) `therefore theta = - (1)/(mu) ((dmu)/(dy))` d `rArr` Above equation gives required result . |
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