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

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,

MARKS App · Accepted Answer

Two refractions will take place, first from spherical surface and the other from the plane surface. So, applying $ v μ 2 ​ ​ − u μ 1 ​ ​ = R μ 2 ​ − μ 1 ​ ​ tw o t im es w i t h p ro p ers i g n co n v e n t i o n . R a yo f l i g h t i s t r a v e ll in g d o w n w a r d s . T h ere f ore , d o w n w a r dd i rec t i o ni s t ak e na s p os i t i v e d i rec t i o n . v 7/4 ​ − − 24 1.0 ​ = + 6 7/4 − 1.0 ​ ( 18 − x ) 4/3 ​ − v 7/4 ​ = ∝ 4/3 − 7/4 ​ ​ S o l v in g t h esee q u a t i o n s , w e g e t x = 2 cm 	herefore A n s w er i s 2. A na l ys i so f Q u es t i o n ( i ) Q u es t i o ni s m o d er a t e l y d i ff i c u lt f ro m c a l c u l a t i o n p o in t o f v i e w , o t h er w i se i t i ss im pl e . ( ii ) \frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R} c anb e a ppl i e df or pl an es u r f a ce a l so w i t h R=\propto$

Search any question & find its solution

Looking for more such questions to practice?