InterviewSolution
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InterviewSolution
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Water (with refractive index = 4/3) in a tank is `18 cm` deep. Oil of refraction index `7//4` lies on water making a convex surface of radius of curvature `R = 6 cm` as shown in Fig. 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. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B For refraction at air-oil interface, we have `u = -24 cm, mu_1 = 1, mu_2 = (7)/(4)`, `R = + 6 cm, v = v_1`(say) As `-(mu_1)/(u)+(mu_2)/(v)=(mu_2 - mu_1)/( R)` `:. -(1)/(-24)+(7//4)/(v_1)=((7//4) - 1)/(6) = (3)/(24)` or `(7)/(4 v_1)=(3)/(24)-(1)/(24)=(2)/(24)=(1)/(12)` or `v_1 = (12 xx 7)/(4) = 21 cm` This image will act as object for oil water interface. Taking refraction at oil-water interface, we have `u = + 21 cm, v = v_2, mu_1 = (7)/(4), mu_2 = (4)/(3), R = oo` As `-(mu_1)/(u)+(mu_2)/(v) = (mu_2 - mu_1)/(R )` `:. (-(7//4))/(21) +((4//3))/(v_2)=(4//3 - 7//4)/(oo) = 0` or `v_2 = 16 cm` Hence, `x = 18 - 16 = 2 cm`. |
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