Question

Two identical thin plano-convex glass lenses of refractive index $1.50$ each having radius of curvature of $20\text{cm}$ are placed with their convex surfaces in contact at the centre. The intervening space is filled with the liquid of refractive index $1.7$ as shown in figure. The focal length of this combination is

• A. $50\text{cm}$
• B. $-20\text{cm}$
• C. $-25\text{cm}$
• D. $-50\text{cm}$

Answer 1

The correct option is D $-50\text{cm}$

If two or more lenses are in contact, the equivalent focal length is given by the formula
$$\frac{1}{f}=\sum _{i=1}^n\frac{1}{f_i}$$
From the data given in the question, we can write that,
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

Focal length of plano-convex lens is given by
$$f=\frac{R}{\mu -1}$$
Substituting the data given in the question we get,
$f=\frac{20}{1.5-1}=40\text{cm}$
$\therefore f_1=40\text{cm},f_2=40\text{cm}$

Since, liquid in the gap behaves as a lens, using lens maker's formula,

$\frac{1}{f_s}=(1.7-1)(-\frac{1}{20}-\frac{1}{20})$
$\frac{1}{f_3}=0.7\times \frac{-1}{10}=\frac{-7}{100}$

Substituting the data obtained we get,

$\frac{1}{f}=\frac{1}{40}+\frac{1}{40}-\frac{7}{100}$
$\Rightarrow \frac{1}{f}=\frac{-2}{100}\Rightarrow f=-50\text{cm}$

Hence, option (d) is the correct answer.