Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7 . The focal length of the combination is

 Given combination is equivalent to three lenses. In which two are plano-convex with refractive index 1.5 and one is concave lens of refractive index 1.7 .

Using lens maker formula, 

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

For plano-convex lens 
$R_1=\infty ,R_2=-20 \mathrm{cm}$

$\therefore \frac{1}{f_1}=\frac{1}{f_2}=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{-20}\right)=\frac{0.5}{20}=\frac{1}{40}$

$\text{ So, }{f}_1=f_2=40 \mathrm{cm}$

For concave lens, 

$\mu =1.7,R_1=-20 \mathrm{cm},R_2=20 \mathrm{cm}$

$\therefore \frac{1}{f_3}=(1.7-1)\left(\frac{1}{-20}-\frac{1}{20}\right)$

$=0.7\times \left(\frac{-2}{20}\right)=-\frac{7}{100}$

$\text{ So, }{f}_3=-\frac{100}{7} \mathrm{cm}$

$\text{ Equivalent focal length }\left(f_{eq}\right)\text{ of the system is given by }$

$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_3}+\frac{1}{f_2}=\frac{1}{40}+\frac{1}{-100/7}+\frac{1}{40}$

$=\frac{1}{20}-\frac{7}{100}=-\frac{2}{100}=-\frac{1}{50}$

$\therefore f_{eq}=-50 \mathrm{cm}$