Question

# Water (with refractive index $$=\displaystyle \dfrac{4}{3}$$) in a tank is $$18\ cm$$ deep. Oil of refractive index $$\displaystyle \dfrac{7}{4}$$ 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. Then '$$x$$' is :

A
2
B
3
C
4
D
5

Solution

## 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)$$\mu_1=1, \mu_2=\dfrac{7}{4}, u=-24 cm, R=6 cm$$Putting this into the formula, we get$$\dfrac {\dfrac{7}{4}}{v}-\dfrac {1}{-24}=\dfrac {\dfrac{7}{4}-1}{6}$$Solving this, we get $$v=21 cm$$As the lens is thin, this can be taken as the object distance for the second refraction.For the second refraction,(by sign convention)$$\mu_1=\dfrac{7}{4}, \mu_2=\dfrac{4}{3}, u=21 cm, R=\infty$$ (plane surface)Putting this into the formula, we get$$\dfrac {\dfrac{4}{3}}{v}-\dfrac {\dfrac{7}{4}}{21}=0$$Solving this, we get $$v=16 cm$$ i.e. it is 16 cm below the water level.The tank is $$18 cm$$ deep.So the height of the image from the bottom of the tank is $$18-16= 2 cm$$PhysicsNCERTStandard XII

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