JEE MAIN - Physics (2014 (Offline) - No. 9)
A thin convex lens made from crown glass $$\left( {\mu = {3 \over 2}} \right)$$ has focal length $$f$$. When it is measured in two different liquids having refractive indices $${4 \over 3}$$ and $${5 \over 3},$$ it has the focal lengths $${f_1}$$ and $${f_2}$$ respectively. The correct relation between the focal lengths is :
$${f_1} = {f_2} < f$$
$${f_1} > f$$ and $${f_2}$$ becomes negative
$${f_2} > f$$ and $${f_1}$$ becomes negative
$${f_1}\,$$ and$${f_2}\,$$ both become negative
Explanation
By Lens maker's formula for convex lens
$${1 \over f} = \left( {{\mu \over {{\mu _L}}} - 1} \right)\left( {{2 \over R}} \right)$$
for, $$\mu {L_1} = {4 \over 3},{f_1} = 4R$$
for $$\mu {L_2} = {5 \over 3},{f_2} = - 5R$$
$$ \Rightarrow {f_2} = \left( - \right)ve$$
$${1 \over f} = \left( {{\mu \over {{\mu _L}}} - 1} \right)\left( {{2 \over R}} \right)$$
for, $$\mu {L_1} = {4 \over 3},{f_1} = 4R$$
for $$\mu {L_2} = {5 \over 3},{f_2} = - 5R$$
$$ \Rightarrow {f_2} = \left( - \right)ve$$
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