Question

Water (with refractive index = $\frac{4}{3}$) in a tank is 18 cm deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature R = 6 cm as shown. Consider oil to act as a thin lens. An object S is placed 24 cm above water surface. The location of its image is at x cm above the bottom of the tank. Find the value of x.

• A. 2 cm
• B. 1 cm
• C. 1.5 cm
• D. 2.5 cm

Answer 1

The correct option is A 2 cm

refrachve index of water uw $=\frac{4}{3}$

refrachve index of oil is $=\frac{7}{4}$

object distance $=u$,

image distance $=v$

$x$ is the location of image

$\text{Now,}\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

downward is taken as +ve direction

i) $\frac{7/4}{v}-\frac{1.0}{-24}=\frac{7/4-1}{+6}$

$\begin{array}{c}\Rightarrow \frac{q}{4v}=\frac{3}{24}-\frac{1}{24}\\ \Rightarrow v=+21cm\end{array}$

Now water act as plane mirron

$\therefore R=\alpha$

$\begin{array}{cc}\text{Hence}& \frac{{\mu }_2}{V}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\\ \therefore & \frac{4/3}{18-1c}-\frac{7/4}{v}=\frac{{\mu }_2-{\mu }_1}{\infty }\\ \therefore & \frac{4}{3(12-t)}-\frac{7/4}{21}=0\end{array}$

$\begin{array}{cc}& \frac{4}{3(18-c)}=\frac{1}{12}\\ \Rightarrow & 18-k=\frac{12\times 4}{3}\\ & =2cm\\ & \text{Hence option (A) is correct}\end{array}$