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A ray is incident at a small angle theta on a glass slab of thickness t. If the refractive index of glass is mu , show that the lateral displacement of the emergent ray from the slab is t theta(mu - 1) //mu. |
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Answer» SOLUTION :Lateral DISPLACEMENT of the emergent ray BD relative to the INCIDENT ray AO is BC [Fig. 2.75]. `therefore "" BC = OB*(BC)/(OB) = OP*(OB)/(OP)*(BC)/(OB)` `= OP sec angleBOP * SIN angleBOC` `= t sec theta. sin(theta - theta.)` `theta and theta. "being small we have",` `sce theta. = (1)/(costheta) approx (1)/(1) = 1 and sin(theta - theta.) approx theta - theta.` `therefore "" BC = t(theta - theta.) = t theta(1 - (theta.)/(theta))` `"again," mu = (sintheta)/(sintheta.) approx = (theta)/(theta.) or, (1)/(mu) = (theta.)/(theta)` `therefore "" "Lateral displacement,"` `BC = t theta(1 - (1)/(mu)) = t theta((mu - 1))/(mu)` 
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