Question

Water (with refractive index $=\frac{4}{3}$) in a tank is $18cm$ 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 $24cm$ above water surface. The location of its image is at '$x$' cm above the bottom of the tank. Then '$x$' is :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

Here, the usual lens equation wont be applicable as the medium is different on two sides of the lens.

So we will consider refraction at two surfaces.

For first refraction, (by sign convention)

${\mu }_1=1,{\mu }_2=\frac{7}{4},u=-24cm,R=6cm$

Putting this into the formula, we get

$\frac{\frac{7}{4}}{v}-\frac{1}{-24}=\frac{\frac{7}{4}-1}{6}$

Solving this, we get $v=21cm$

As the lens is thin, this can be taken as the object distance for the second refraction.

For the second refraction,(by sign convention)

${\mu }_1=\frac{7}{4},{\mu }_2=\frac{4}{3},u=21cm,R=\infty$ (plane surface)
Putting this into the formula, we get

$\frac{\frac{4}{3}}{v}-\frac{\frac{7}{4}}{21}=0$

Solving this, we get $v=16cm$ i.e. it is 16 cm below the water level.
The tank is $18cm$ deep.

So the height of the image from the bottom of the tank is $18-16=2cm$