The power of a lens (biconvex) is $1.25 \mathrm{~m}^{-1}$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $20 \mathrm{~cm}$ and $40 \mathrm{~cm}$ respectively. The refractive index of surrounding medium:
Solution
<p>$\because$ $${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$</p>
<p>$$ \Rightarrow {{1.25} \over {100}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {20}} + {1 \over {40}}} \right)$$</p>
<p>$$ \Rightarrow {1 \over {80}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right) \times {{(4 + 2)} \over {80}}$$</p>
<p>$\Rightarrow {{1.5} \over {{\mu _1}}} - 1 = {1 \over 6}$</p>
<p>$\Rightarrow {{1.5} \over {{\mu _1}}} = {7 \over 6}$</p>
<p>$\Rightarrow {\mu _1} = {{1.5 \times 6} \over 7} = {9 \over 7}$</p>
About this question
Subject: Physics · Chapter: Optics · Topic: Lenses and Optical Instruments
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