JEE Physics Solution: Question 12
Logical Approach
To determine the “radius of the disc visible,” we must identify the path of light rays that allow an observer at a “large height” to see points on the disc. The condition “large height” implies that the observer only captures rays that emerge from the top base of the cone parallel to the vertical axis.
We will trace such a ray backwards from the observer’s eye to the disc:
- Entry from Top: A ray traveling vertically downwards from the observer enters the flat, horizontal base of the cone normally. Thus, it continues vertically downwards inside the glass.
- Interaction with Side Wall: The vertical ray strikes the slanted side of the cone. We must check if it refracts out or undergoes Total Internal Reflection (TIR).
- Path to Disc: Based on the reflection, we trace the ray until it intersects the plane of the disc.
Fig 1: Ray tracing relating the visible radius on the disc to the aperture on the top surface.
Step-by-Step Derivation
1. Ray Geometry inside the Cone
Consider a ray entering the top horizontal surface at a distance $x$ from the vertical axis. Since the observer is at a large height, the incident ray is vertical.
- Refraction at Top Surface: The ray enters normally (angle of incidence $0^\circ$) and proceeds vertically downwards.
- Incidence at Side Wall: The cone has a semi-vertical angle $\alpha = 30^\circ$. The angle between the vertical ray and the slanted surface is $30^\circ$. Therefore, the angle of incidence with the normal is $i = 90^\circ – 30^\circ = 60^\circ$.
2. Total Internal Reflection (TIR) Check
Using Snell’s Law, we calculate the critical angle $C$ for the glass ($\mu = 1.5$) to air interface:
$$ \sin C = \frac{1}{\mu} = \frac{1}{1.5} \approx 0.667 \implies C \approx 41.8^\circ $$Since the angle of incidence $i = 60^\circ > C$, Total Internal Reflection occurs.
3. Path after Reflection
The ray reflects with an angle of reflection $r = 60^\circ$. Geometrically, the reflected ray travels perpendicular to the opposite slanted face.
4. Exit and Interception with the Disc (Geometric Proof)
Since the ray strikes the second face normally ($90^\circ$), it passes through undeviated to the disc. We now mathematically derive the relationship between the entry coordinate $x_{top}$ and the disc coordinate $r_{disc}$.
Let us set up a Cartesian coordinate system where the origin $(0,0)$ is at the tip of the cone (center of the disc). The $y$-axis is the vertical axis of the cone.
- Equation of the Cone Surface: Since the semi-vertical angle is $30^\circ$, the slanted face makes an angle of $60^\circ$ with the horizontal ($x$-axis). The equation of the right face is: $$ y = (\tan 60^\circ)x = \sqrt{3}x $$
- Incident Point ($P$): A vertical ray at distance $x_{top}$ travels downwards and intersects this face. Substituting $x = x_{top}$ into the line equation, the intersection point $P$ is $(x_{top}, \sqrt{3}x_{top})$.
- Reflected Ray Equation: The incident ray reflects off the surface. Due to the $60^\circ$ incidence and reflection, the reflected ray travels perpendicular to the opposite face (the left face, which has slope $-\sqrt{3}$).
Therefore, the slope of the reflected ray is $m = \frac{1}{\sqrt{3}}$.
Using the point-slope form $y – y_1 = m(x – x_1)$ at point $P$: $$ y – \sqrt{3}x_{top} = \frac{1}{\sqrt{3}}(x – x_{top}) $$ - Intersection with Disc ($y=0$): The disc lies on the horizontal plane passing through the tip. Setting $y=0$: $$ 0 – \sqrt{3}x_{top} = \frac{1}{\sqrt{3}}(x – x_{top}) $$ Multiply both sides by $\sqrt{3}$: $$ -3x_{top} = x – x_{top} $$ $$ x = -2x_{top} $$
The negative sign indicates the ray crosses to the opposite side of the axis. Taking the magnitude, the radius on the disc is twice the radius on the top surface:
$$ r_{\text{disc}} = 2 \cdot x_{\text{top}} $$The question asks for the visible radius of the disc. The visible region corresponds to the aperture $x_{\text{top}}$ for the full disc radius $R_{\text{disc}} = 4.0 \text{ cm}$.
$$ x_{\text{top}} = \frac{R_{\text{disc}}}{2} $$ $$ x_{\text{top}} = \frac{4.0 \text{ cm}}{2} = 2.0 \text{ cm} $$