Question

An equiconvex lens of refractive index $\frac{3}{2}$ and focal length 10 cm is held with its axis vertical and its lower surface immersed in water($\mu$=$\frac{4}{3}$, the upper surface being in air.At wat distance from the lens ,will a vertical beam of parallet light incident on the lens be focused?

• A. 20 cm
• B. 30 cm
• C. 10 cm
• D. 5 cm

Answer 1

The correct option is A 20 cm

$\begin{array}{c}\text{According to lens-maxers formula,}\\ \begin{array}{c}\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})\\ \text{Here,}f=10cm,\mu =\left(3/2\right),R_1=R\\ R_2=-R\end{array}\\ \begin{array}{}\end{array}\end{array}$

$\begin{array}{c}\text{So,}\frac{1}{10}=(\frac{3}{2}-1)(\frac{1}{R}+\frac{1}{R})\\ R=10cm\text{Now, in case of refraction from}\\ \text{curred surface,}\\ \frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\\ \text{So, for air glass boundary}\end{array}$

$\begin{array}{c}\frac{\left(3/2\right)}{V_A}-\frac{1}{-\infty }=\frac{\left(3/2\right)-1}{+10}\\ \text{i.e,}{V}_A=30cm\\ \text{i.e, the air glass surface forms}\\ \text{image.}A\text{at a distance of}30cm\\ \text{from it as shown in fig. This}\\ \text{image will act as an object for}\end{array}$

$\begin{array}{c}\text{glass - water interface which is}\\ \text{curved, so that}\\ \begin{array}{cc}\frac{\left(4/3\right)}{v}& -\frac{\left(3/2\right)}{+30}=\frac{\left(4/3\right)-\left(3/2\right)}{-10}\\ \text{i.e,}& \frac{4}{3v}=\frac{1}{60}+\frac{1}{20}\\ \Rightarrow v=20cm& \end{array}\end{array}$

$\begin{array}{c}\text{So, the incident parallel beam}\\ \text{will converge at a distance}\\ \text{of}20cm\text{from the lens}\\ \text{inside the water.}\end{array}$