JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 15)
A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid
of refractive index 1.42. If its focal length in liquid is f1
, then the ratio $${{{f_1}} \over f}$$ is closest to the
integer :
17
1
9
5
Explanation
Using formula
$${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$${1 \over {{f}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ ...(1)
and $${1 \over {{f_1}}} = \left( {{{1.5} \over {1.42}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ ...(2)
Dividing (1) by (2), we get
$${{{f_1}} \over f} = {{0.5} \over {0.056}}$$ = 8.93 $$ \approx $$ 9
$${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
$${1 \over {{f}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ ...(1)
and $${1 \over {{f_1}}} = \left( {{{1.5} \over {1.42}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$ ...(2)
Dividing (1) by (2), we get
$${{{f_1}} \over f} = {{0.5} \over {0.056}}$$ = 8.93 $$ \approx $$ 9
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