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