Water (refractive index = 4/3) in a tank is 19 cm deep. Oil of refractive index 7/4 lying on water makes a convex surface of radius of curvature 6 cm and acts as a thin lens. An object S is placed 24 cm above the water surface. Its image is formed at x cm above the bottom of the tank. Then x = _____
Physics · Mixed Chapter
JEE Mains

Question

Water (refractive index = 4/3) in a tank is 19 cm deep. Oil of refractive index 7/4 lying on water makes a convex surface of radius of curvature 6 cm and acts as a thin lens. An object S is placed 24 cm above the water surface. Its image is formed at x cm above the bottom of the tank. Then x = _____

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Treat the curved air–oil boundary as a spherical refracting surface of radius +6 cm (downward as positive); n₁=1, n₂=7/4, and object distance u=−24 cm. Step 2: Use n₂/v−n₁/u=(n₂−n₁)/R. Thus (7/4)/v+1/24=(3/4)/6=1/8, giving (7/4)/v=1/12 and v=21 cm. Step 3: The subsequent plane oil–water interface (refractive index of water=4/3) changes the paraxial image distance to image distance in water=v·((refractive index of water)/(refractive index of oil))=21×(4/3)/(7/4)=16 cm below the water sur…