Water (with refractive index = 3 4 ) in a tank is 18 cm deep. Oil of refractive index 4 7 lies on water making a convex surface of radius of curvature R = 6 cm as shown. Consider oil to act as a t

Question

Water (with refractive index = 3 4 ​ ) in a tank is 18 cm deep. Oil of refractive index 4 7 ​ 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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Two refractions will take place, first from spherical surface and the other from the plane surface. So, applying $\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$ two times with proper sign convention. Ray of light is travelling downwards. Therefore, downward direction is taken as positive direction. $\frac{7/4}{v} - \frac{1.0}{- 24} = \frac{7/4 - 1.0}{+ 6} \frac{4/3}{( 18 - x )} - \frac{7/4}{v} = \frac{4/3 - 7/4}{\propto}$ Solving these equations, we get $x = 2 cm$ $∴$ Answer is 2 . Analysis of Question (i) Question is moderately difficult from calculation point of view, otherwise it is simple. (ii) $\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}$ can be applied for plane surface also with $R =\propto$

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