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An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle with its tip on the principle axis of the lens if moved along the axis intil its inverted real image coincides with the needle itself, Fig. The distance of needle from lens is measured to be a. On removing the liquid layer and repeating the experiment, the distance isn found to be b. Given that two values of distances measured represent that real length values in the two cases, obtain a formula for refractive index of the liquid. .

Answer» <html><body><p></p>Solution :Here, combined focal length of <a href="https://interviewquestions.tuteehub.com/tag/glass-14164" style="font-weight:bold;" target="_blank" title="Click to know more about GLASS">GLASS</a> lens and liquid lens, `F = a`, and Focal length of <a href="https://interviewquestions.tuteehub.com/tag/convex-933348" style="font-weight:bold;" target="_blank" title="Click to know more about CONVEX">CONVEX</a> lens, `f_1 = b` If `f_2` is focal length of liquid lens, then <br/> `(1)/(f_1)+(1)/(f_2)=(1)/(F)` <br/> `(1)/(f_2)=(1)/(F)-(1)/(f_1) = ((1)/(a) - (1)/(b))` <br/> The liquid lens is plano <a href="https://interviewquestions.tuteehub.com/tag/concave-927978" style="font-weight:bold;" target="_blank" title="Click to know more about CONCAVE">CONCAVE</a> lens for which `R_1 = -r, R_2 = oo` <br/> From `(1)/(f_2)=(<a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a> -1) ((1)/(R_1) - (1)/(R_2))` <br/> `((1)/(a) - (1)/(b)) = (mu - 1) ((1)/(-r) - (1)/(- oo))` <br/> `:. (mu - 1) = (r)/(b) - (r)/(a)` <br/> `mu = 1 + (r)/(b) - (r)/(a)`.</body></html>


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