JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Geometrical Optics

A beaker of radius

r

is filled with water (refractive index

\frac{4}{3}

) up to a height

H

as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed

ω

. This makes the water surface curved so that the difference in the height of water level at the center and at the circumference of the beaker is

h (h ≪ H, h ≪ r)

, as shown in the figure on the right. Take this surface to be approximately spherical with a radius of curvature

R

. Which of the following is/are correct? (g is the acceleration due to gravity)

(A) $R = \frac{h ^{2} + r ^{2}}{2 h}$ ✓ Correct
(B) $R = \frac{3 r ^{2}}{2 h}$
(C)Apparent depth of the bottom of the beaker is close to $\frac{3 H}{2} (1 + \frac{ω ^{2} H}{2 g})^{- 1}$
(D)Apparent depth of the bottom of the beaker is close to $\frac{3 H}{4} (1 + \frac{ω ^{2} H}{4 g})^{- 1}$ ✓ Correct

Answer: (A), (D)

Solution

From geometry of the spherical surface, in triangle OAB:

R^{2} = (R - h)^{2} + r^{2}

R^{2} = R^{2} - 2 h R + h^{2} + r^{2}

R = \frac{h ^{2} + r ^{2}}{2 h}

✓ (Option A)

For a rotating liquid surface:

h = \frac{ω ^{2} r ^{2}}{2 g}

Using refraction formula

\frac{μ _{2}}{v} - \frac{μ _{1}}{u} = \frac{μ _{2} - μ _{1}}{R}

\frac{1}{v} + \frac{4}{3 ( H - h )} = \frac{1 - 4/3}{- R}

Substituting

R = \frac{h ^{2} + r ^{2}}{2 h} \approx \frac{r ^{2}}{2 h}

(since

h ≪ r

) and

h = \frac{ω ^{2} r ^{2}}{2 g}

\frac{1}{v} = \frac{2 h}{3 r ^{2}} - \frac{4}{3 H} = - \frac{4}{3 H} [1 - \frac{ω ^{2} H}{4 g}]

v = \frac{3 H}{4} [1 + \frac{ω ^{2} H}{4 g}]^{- 1}

✓ (Option D)

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options