A beaker of radius r is filled with water (refractive index 4/3) up to height H and rotates with angular speed ω. The water surface becomes approximately spherical; the height difference between centre and rim is h (h << H, h << r), and the radius of curvature is R. Which statement(s) is/are correct?

Question

G1 Physics · Accepted Answer

For a spherical surface with sagitta h and aperture radius r, R=(r²+h²)/(2h), so A is correct. For rotating water, the curvature produces an additional refracting power. Using water refractive index 4/3 and the small-curvature approximation, the apparent depth is (3H/4)[1 + Hω²/(4g)]⁻¹. Hence A and D are correct.