Verified Solution • 2025–26 Session moderate

QJEEM22Phe060: JEE Main 2022 Physics - Class 12 JEEM22Phe060 Solution: Two identical thin biconvex lens of focal length 15 cm and…

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.

Step-by-Step Verified Solution

1fl=(μeμm−1)(1R1−1R2){1 \over {{f_l}}} = \left( {{{{\mu _e}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

here ∣R1∣=∣R2∣=R|{R_1}| = |{R_2}| = R

⇒1fl1=(1.5−1)(2R)=115\Rightarrow {1 \over {{f_{{l_1}}}}} = (1.5 - 1)\left( {{2 \over R}} \right) = {1 \over {15}}

⇒1R=115\Rightarrow {1 \over R} = {1 \over {15}} or R=15R = 15 cm

for the concave lens made up of liquid

1fl2=(1.25−1)(−2R)=−130{1 \over {{f_{{l_2}}}}} = (1.25 - 1)\left( { - {2 \over R}} \right) = - {1 \over {30}} cm

now for equivalent lens

1fe=2fl1+1fl2{1 \over {{f_e}}} = {2 \over {{f_{{l_1}}}}} + {1 \over {{f_{{l_2}}}}}

=215−130=330=110= {2 \over {15}} - {1 \over {30}} = {3 \over {30}} = {1 \over {10}}

or fe=10{f_e} = 10 cm

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