Question

Two identical thin plano-convex glass lenses $($refractive index $1.5)$ 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 oil of refractive index $1.7$. The focal length of the 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 $-50\text{cm}$


Given,
refractive index of glass, ${\mu }_g=1.5$
radius of curvature, $R=20\text{cm}$
refractive index of oil, ${\mu }_o=1.7$

Let the focal length of the combination is $f$. So, we can write,

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}.....\left(1\right)$

From lens makers formula for the plano-convex lens,
$\frac{1}{f_1}=({\mu }_g-1)(\frac{1}{R_1}-\frac{1}{R_2})...\left(2\right)$

Here, $R_1=R=20\text{cm}$ and for plane surface, $R_2=\infty$.

Substituting the values in $\left(2\right)$, we have
$\frac{1}{f_1}=(1.5-1)(\frac{1}{20}-\frac{1}{\infty })$

$\therefore ,f_1=40\text{cm}$

Since both the plano-convex combination are same,so
$f_2=f_1=40\text{cm}$

When the intervening medium is filled with oil, then focal length of the concave lens formed by the oil, by lens makers formula,

$\frac{1}{f_3}=(1.7-1)(\frac{1}{-R}-\frac{1}{R})$

$\Rightarrow \frac{1}{f_3}=\left(0.7\right)(\frac{1}{-20}-\frac{1}{20})$

$\Rightarrow \frac{1}{f_3}=\frac{0.7}{-10}$

Now, putting the value of $f_1$, $f_2$ and $f_3$ in equation $\left(1\right)$, we get

$\frac{1}{f}=\frac{1}{40}+\frac{1}{40}+\frac{0.7}{-10}$

$\Rightarrow \frac{1}{f}=\frac{2}{40}-\frac{2.8}{40}$

$\Rightarrow \frac{1}{f}=-\frac{0.8}{40}$

$\therefore f=-50\text{cm}$

Thus, option $\left(d\right)$ is the correct answer.