Question

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 :

• A. 50 cm
• B. -20 cm
• C. -25 cm
• D. -50 cm

Answer 1

The correct option is

D

-50 cm

We know,

Focal length of combination is given as

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$------(1)

For the first part, refractive index is 1.5
$\frac{1}{f_1}=(1.5-1)(\frac{1}{\infty }-\frac{1}{-20})=\frac{1}{40}$

For the second part, refractive index is 1.5

$\frac{1}{f_3}=(1.5-1)(\frac{1}{-20}-\frac{1}{\infty })=\frac{1}{40}$

For the part in which oil is filled, refractive index is 1.7
$\frac{1}{f_2}=(1.7-1)(\frac{1}{-20}-\frac{1}{20});\frac{1}{f_2}=(1.7-1)(\frac{1}{-20}-\frac{1}{20})=-\frac{0.7\times 2}{20}=-\frac{2.8}{40}$

Now putting all the values in equation 1
$\frac{1}{f}=\frac{1}{40}-\frac{2.8}{40}+\frac{1}{40}=\frac{1-2.8+1}{40}=-\frac{0.8}{40}$
$f=-\frac{40}{0.8}=-50cm$

Hence, resultant focal length is -50 cm