Question

A small bulb is placed at the bottom of a tank, containing a transparent liquid of refractive index \(\sqrt{2}\), to a depth of \(1\,\text{m}\). Calculate the area of the surface of the liquid through which light from the bulb emerges.

Hint:

For such total internal reflection problems, \[ r=h\tan C \] and \[ A=\pi r^2. \] Always calculate the critical angle first and then determine the radius of the illuminated circular region.

Answer 1

Concept: When a source of light is placed inside a denser medium, light rays incident at the liquid-air interface with angle of incidence less than or equal to the critical angle emerge into air. Rays incident at angles greater than the critical angle suffer total internal reflection and do not emerge. Therefore, only those rays lying inside a cone of semi-vertical angle equal to the critical angle can come out of the liquid surface. The intersection of this cone with the liquid surface forms a circular region through which light emerges. The required area is therefore the area of this circular patch.

Step 1: Determine the critical angle for the liquid-air interface. For a denser medium of refractive index \(n\) and air of refractive index \(1\), \[ \sin C=\frac{1}{n}. \] Given, \[ n=\sqrt{2}. \] Therefore, \[ \sin C=\frac{1}{\sqrt{2}}. \] Since \[ \sin 45^\circ=\frac{1}{\sqrt{2}}, \] we obtain \[ C=45^\circ. \] Thus, the critical angle of the liquid-air interface is \[ 45^\circ. \]

Step 2: Determine the radius of the circular region. Let \[ h=1\,\text{m} \] be the depth of the bulb. The extreme ray that can emerge from the liquid strikes the surface at the critical angle. From the right triangle formed, \[ \tan C=\frac{r}{h}, \] where \(r\) is the radius of the circular patch. Substituting values, \[ \tan45^\circ=\frac{r}{1}. \] Since \[ \tan45^\circ=1, \] we obtain \[ r=1\,\text{m}. \]

Step 3: Calculate the area of the circular patch. Area through which light emerges is \[ A=\pi r^2. \] Substituting \(r=1\,\text{m}\), \[ A=\pi(1)^2. \] \[ A=\pi\,\text{m}^2. \] Hence, \[ A\approx3.14\,\text{m}^2. \]

Step 4: Write the final answer. Therefore, the area of the liquid surface through which light emerges is \[ \boxed{\pi\,\text{m}^2} \] or \[ \boxed{3.14\,\text{m}^2}. \]