QJEEM25Phc36e: JEE Main 2025 Physics - Class 12 JEEM25Phc36e Solution: Two identical symmetric double convex lenses of focal length f…

Here,

we will use lens maker formula,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

So, the focal length of L $_1$ remains the same as the original lens because the curvature of the surfaces does not change

So, ${\left( f \right)_{{L_1}}} = f$ .... (1)

Now, for L $_3$ , L $_3$ is plano-convex lens

${R_1} = R,\,{R_2} = \infty$

So, ${1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over \infty }} \right)$

$\Rightarrow {1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right){1 \over R}$ ..... (2)

For whole double convex lens,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right)$ (As ${R_1} = R$ , ${R_2} = - R$ )

$\Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} + {1 \over R}} \right)$

$\Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{2 \over R}} \right)$ .... (3)

From (2) and (3),

${1 \over {{{\left( f \right)}_{{L_3}}}}} = {1 \over {2f}} \Rightarrow {\left( f \right)_{{L_3}}} = 2f$

Hence, ${{{{\left( f \right)}_{{L_1}}}} \over {{{\left( f \right)}_{{L_3}}}}} = {f \over {2f}} \quad\text{....(from (1)})\quad= 1:2$