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. Then 'x' is :

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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. Then 'x' is :

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$\frac{μ_{3}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R_{1}} + \frac{μ_{3} - μ_{2}}{R_{2}}$

$\frac{4}{3 V} - \frac{1}{- 24} = \frac{\frac{7}{4} - 1}{6} + \frac{\frac{4}{3} - \frac{7}{4}}{\infty} \Rightarrow \frac{4}{3 V} + \frac{1}{24} = \frac{1}{8} \Rightarrow \frac{4}{3 V} = \frac{1}{12} \Rightarrow V = 16 cm$

$∴$ Ans. = (18 –16) cm = 2 cm

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