JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 16)
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:
1.0
$$\frac{9}{7}$$
$$\frac{3}{2}$$
$$\frac{4}{3}$$
Explanation
$$\because$$ $${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$$ \Rightarrow {{1.25} \over {100}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {20}} + {1 \over {40}}} \right)$$
$$ \Rightarrow {1 \over {80}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right) \times {{(4 + 2)} \over {80}}$$
$$ \Rightarrow {{1.5} \over {{\mu _1}}} - 1 = {1 \over 6}$$
$$ \Rightarrow {{1.5} \over {{\mu _1}}} = {7 \over 6}$$
$$ \Rightarrow {\mu _1} = {{1.5 \times 6} \over 7} = {9 \over 7}$$
Comments (0)
