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A cicular disc of radius Ris placed co-axially and horizontally inside an opaque hemi spherical bowl of radius .a. (see figure). The faredge of the disc in just visible when viewed from the edge of the bowl . The bowl is filled with transparent liquid of refractive index mu and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed ? |
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Answer» Solution :CONSIDER situartion before pouring water in an opaque bowl of radius a. Now consider a CIRCULAR disc of radius R and centre C which si placed inside the bowl horizontnally and coaxially with the bowl. Here we have to calculate OC=d . `rArr` Here before filling bowl with water, incident light ray is AMA. . Now when water is paured in the hemispherical bowl completely, the nearer and B of the disc is seen for which inciden ray is `vec (BM)` and refracted ray is `vec(MA)`. Here NN. is perpedicular to water surface, drawn at point M. Hence `anlge (BMN)` = i (angle of incidence) and `angle NMA = alpha` (angle of refraction) `rArr` Now, applying Snell.slaw at point M , `u sin i = (1) sin r` `thereforemu sin i = sin alpha (therefore r = alpha)` `therefore 1/mu = (sin i)/(sin alpha)` ........(1) `[because r = alpha]` Now, from , figure `sin i= (BN.)/(BM)` but `BN.= CN. - CB = OM -CB` and `BM = SQRT(d^2 + (a-R)^2)` `therefore sin i = sqrt(a-R)/(sqrt(d^2 + (a-R)^2))`...... (2) and `sin alpha = cos(90^@ - alpha) = (AN.)/(AM) = (AC + CN.)/(AM)` `thereforesin alpha = (AC +OM)/(AM)` `= (a+R)/(sqrt(d^2(a+R)^2)) ........(3)` `therefore` From equation (1),(2) and (3), `1/mu= (a-R)/(sqrt(d^2 +(a-R)^2)) xx (sqrt(d^2 + (a+R)^2))/(a+R)` `therefore mu(a-R)(sqrt(d^2 + (a+R)^2))=(a+R)(sqrt(d^2+(a-R)^2))` Taking squares on the both sides, `therefore mu^2(a-R)^2 {d^2 + (a+R)^2}=(a+R)^2{d^2 + (a-R)^2}` `therefore {mu^2(a-R)^2d^2} + {mu^2(a-R)^2(a+R)^2} = {d^2(a+R)^2} + {(a+R)^2(a-R)^2}` `therefore(a-R)^2(a+R)^2(mu^2 -1)=d^2{(a+R)^2-mu^2(a-R)^2}` `therefore d^2 = ((mu^2-1){(a-R)(a+R)}^2)/((a+R)^2 - mu^2 (a-R)^2)` `therefore d^2 = ((mu^2-1)(a^2-R^2)^2)/((a+R)^2-mu^2(a-R)^2)` `therefore d=sqrt((mu^2-1)/((a+R)^2-mu^2(a-R)^2))xx (a^2 xxb^2)` `rArr`Above equation gives required result. |
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