An equi-convex 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 principal axis of the lens is moved along the axis until its inverted real image coincides with the needle itself. The distance of needle from lens is measured to be a. On removing the liquid layer and repeating the experiment, the distance is found to b. Given that two values of distances measured represent the real length values in the two cases, obtain a formula for refractive index of the liquid. - Zigya
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An equi-convex 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 principal axis of the lens is moved along the axis until its inverted real image coincides with the needle itself. The distance of needle from lens is measured to be a. On removing the liquid layer and repeating the experiment, the distance is found to b. Given that two values of distances measured represent the real length values in the two cases, obtain a formula for refractive index of the liquid.


Here, combined focal length of glass lens and liquid lens, F = a, and Focal length of convex lens, f1 = b.
If f2 is focal length of liquid lens, then

Here, combined focal length of glass lens and liquid lens, F = a, and
1 over straight f subscript 1 plus 1 over straight f subscript 2 space equals space 1 over straight F
space space space space space space space space space space space space 1 over straight f subscript 2 space equals 1 over straight F minus 1 over straight f subscript 1 space equals space 1 over straight a minus 1 over straight b
The liquid lens is plano-concave lens for which
                           straight R subscript 1 space equals space minus straight r comma space space space space straight R subscript 2 space equals space infinity
From               1 over straight f subscript 2 space equals space left parenthesis straight mu minus 1 right parenthesis space open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses
                    1 over straight a minus 1 over straight b space equals space left parenthesis straight mu minus 1 right parenthesis space open parentheses fraction numerator 1 over denominator negative straight r end fraction minus 1 over infinity close parentheses
therefore          open parentheses straight mu minus 1 close parentheses space equals space straight r over straight b minus straight r over straight a
                       straight mu space equals 1 plus straight r over straight b minus straight r over straight a.
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