Q.A small bulb is placed at the bottom of a tank containing water to a depth of $80\ \text{cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

Light escapes only through the circular patch of surface directly above the bulb, bounded by the critical angle $i_c$ for water-air ($\sin i_c = 1/n$); beyond $i_c$, total internal reflection keeps the light inside.

1. $\sin i_c = 1/1.33 \approx 0.7519 \Rightarrow i_c\approx48.75^\circ$.
2. Radius of the escaping circle: $r = h\tan i_c = 80\times1.140 \approx 91.2\ \text{cm}$ (depth $h=80\ \text{cm}$). …

Why only a circular patch lets light out

Light travelling from water (denser, $n=1.33$) to air (rarer, $n=1$) bends away from the normal. Beyond a certain critical angle $i_c$, the refracted ray would have to bend more than $90^\circ$ from the normal - which is impossible - so instead the light undergoes total internal reflection and never leaves the water. Only rays that strike the surface at angles up to $i_c$ actually emerge.

From a point source at the bottom, rays spread out in every direction; the ones that manage to escape trace out a cone (apex at the bulb, half-angle $i_c$) whose base is a circle on the water's surface, directly above the source.

Step 1: find the critical angle

Step 2: relate the radius of the circle to the depth

The ray that just grazes the critical angle traces the edge of the escaping cone. In the right triangle formed by the bulb, the point directly above it, and the edge of the circle on the surface:

Step 3: compute the area …

Method: Critical Angle & Cone of Emergence

This problem uses the concept of total internal reflection at a plane surface. Light from a point source at the bottom can only escape through a circular area on the water surface — outside this circle, the angle of incidence exceeds the critical angle and light is reflected back.

Steps

Step 1: Find the critical angle for water-air interface

The critical angle $i_c$ is given by:

So:

Step 2: Relate the critical angle to the geometry

Draw a ray from the bulb at the bottom that just grazes the water surface at the critical angle. This ray reaches the surface at a point at distance $r$ from the vertical line above the bulb.

From the right triangle formed:

• Depth of water = $h = 80\ \text{cm}$
• Radius of the circle on the surface = $r$
• Angle at the bulb = $i_c$

We have:

Step 3: Calculate $r$

First compute $\sin i_c$:

Then:

Now:

Therefore: …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Real Depth with Apparent Depth

The error: Students often treat the bulb's actual depth ($80\ \text{cm}$) as the object distance for the refraction formula directly, without considering that the image formed by refraction is virtual and at a different location.

Why it's wrong: For a point source at the bottom, the rays emerging into air appear to come from a virtual image above the actual bulb. The critical angle condition depends on the real depth, not the apparent depth.

How to avoid: Always draw the ray diagram. The bulb is at real depth $h = 80\ \text{cm}$. The critical angle $\theta_c$ is determined by Snell's law at the water-air interface:

The radius $r$ of the circular patch on the water surface is:

Key: Use real depth $h$, not apparent depth.

Mistake 2: Using $\sin\theta_c = n_2/n_1$ Incorrectly

The error: Writing $\sin\theta_c = \frac{n_{\text{water}}}{n_{\text{air}}}$ instead of $\frac{n_{\text{air}}}{n_{\text{water}}}$.

Why it's wrong: For total internal reflection, light travels from denser (water) to rarer (air) medium. The critical angle formula is:

How to avoid: Always identify which medium light is leaving (denser) and which it is entering (rarer). The smaller refractive index goes in the numerator.

Mistake 3: Forgetting the Circular Geometry

The error: After finding $\theta_c$, students sometimes use $r = h \sin\theta_c$ or $r = h / \tan\theta_c$.

Why it's wrong: From the geometry (right triangle with height $h$ and base $r$):

How to avoid: Draw the triangle: vertical side = depth $h$, horizontal side = radius $r$, angle at the bulb = $\theta_c$. Then apply $\tan$.

Mistake 4: Calculating Area Incorrectly

The error: Using $A = \pi r$ or $A = 2\pi r$ instead of $A = \pi r^2$. …