Question

A water drop is laced on the glass plate A convex lens of radii of curvature 20, 20 cm $\mu =1.5$ is placed on it.The focal length of water lens so formed $(\mu ofwater$4/3)

• A. 20 cm
• B. 60 cm
• C. -60 cm
• D. 100 cm

Answer 1

The correct option is C -60 cm

$\begin{array}{c}{\mu }_{\omega }=\frac{4}{3}\\ R_1\left(\text{lens}\right)=-20cm\text{(-ve, it is}{2}^{nd}\text{sarface)}\\ R_2\left(\text{mirror}\right)=\infty \end{array}$

We know that

$\begin{array}{c}\frac{1}{f}=({\mu }_w-1)(\frac{1}{R_1}-\frac{1}{R_2})\\ \Rightarrow \frac{1}{f}=(\frac{4}{3}-1)(\frac{1}{-20}-\frac{1}{\alpha })\\ \Rightarrow \frac{1}{f}=\frac{1}{3}\times \frac{1}{-20}\\ \Rightarrow f=-60cm\end{array}$

Hence, (c) is correct option