Question

A tank is filled with a liquid of refractive index \( \sqrt{2} \), up to a height of 30 cm. A tiny bulb is glowing at the bottom of the tank. Calculate the diameter of an opaque disc floating symmetrically on the liquid surface that can cut off completely the light from the bulb that comes out of the liquid surface.

Hint:

The critical angle for total internal reflection depends on the refractive index of the medium. When the angle of incidence exceeds this critical angle, the light is totally reflected within the medium. In this case, the disc must be wide enough to block all light coming from the bulb.

Answer 1

Given:

• Liquid refractive index: \( n_1 = \sqrt{2} \)
• Air refractive index: \( n_2 = 1 \)
• Liquid height: \( h = 30 \, \text{cm} \)

Light from the bulb refracts at the liquid surface. For the disc to block the light, total internal reflection must occur, requiring the angle of incidence to exceed the critical angle. The critical angle \( \theta_c \) is defined by: \[ \sin \theta_c = \frac{n_2}{n_1} \] Substituting the given values: \[ \sin \theta_c = \frac{1}{\sqrt{2}} \] This yields: \[ \theta_c = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45^\circ \] To block all light, the disc must cover all rays exiting the bulb. These rays refract at the liquid surface at an angle of \( \theta_c \). With the bulb 30 cm from the surface, the light spreads in a circle of radius \( r \). The geometry dictates: \[ \tan \theta_c = \frac{r}{h} \] where \( r \) is the radius of the circle and \( h = 30 \, \text{cm} \) is the bulb's distance from the surface. Since \( \tan 45^\circ = 1 \): \[ 1 = \frac{r}{30} \] Therefore, the radius is: \[ r = 30 \, \text{cm} \] The disc's radius is \( 30 \, \text{cm} \), making its diameter: \[ d = 2r = 2 \times 30 = 60 \, \text{cm} \]