JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 20)
Two identical thin biconvex lens of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is __________ cm.
Answer
10
Explanation
$${1 \over {{f_l}}} = \left( {{{{\mu _e}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$$
here $$|{R_1}| = |{R_2}| = R$$
$$ \Rightarrow {1 \over {{f_{{l_1}}}}} = (1.5 - 1)\left( {{2 \over R}} \right) = {1 \over {15}}$$
$$ \Rightarrow {1 \over R} = {1 \over {15}}$$ or $$R = 15$$ cm
for the concave lens made up of liquid
$${1 \over {{f_{{l_2}}}}} = (1.25 - 1)\left( { - {2 \over R}} \right) = - {1 \over {30}}$$ cm
now for equivalent lens
$${1 \over {{f_e}}} = {2 \over {{f_{{l_1}}}}} + {1 \over {{f_{{l_2}}}}}$$
$$ = {2 \over {15}} - {1 \over {30}} = {3 \over {30}} = {1 \over {10}}$$
or $${f_e} = 10$$ cm
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