Solution to Question 7
Physical Analysis:
1. Geometric Setup
- Let the center of the ball be at $x = d = 11 \text{ cm}$ from the wall (origin at wall $x=0$).
- The incident ray is horizontal at a height $y = 4 \text{ cm}$ (radius of the beam).
- Let $R$ be the radius of the ball.
- The ray strikes the ball at a point $P$. The normal at $P$ makes an angle $\theta$ with the horizontal axis.
From the geometry of the sphere:
$$ \sin \theta = \frac{y}{R} = \frac{4}{R} \quad \text{…(1)} $$The horizontal coordinate of point $P$ relative to the ball’s center is $R \cos \theta$. Thus, the horizontal distance from the wall to point $P$ is:
$$ L = d + R \cos \theta = 11 + R \cos \theta $$2. Reflection Analysis
The incident ray is parallel to the axis. By the law of reflection, the angle of incidence is $\theta$ (angle with the normal). The reflected ray makes an angle $\theta$ with the normal.
The total deviation of the ray from the horizontal axis is $\delta = 2\theta$. The reflected ray travels backwards (towards the wall) and upwards (away from the axis).
The slope of the reflected ray is $\tan(2\theta)$. Considering the triangle formed by the reflected ray, the horizontal distance $L$ and the vertical rise $\Delta y$:
$$ \Delta y = Y_{shadow} – y_{incident} = 52 – 4 = 48 \text{ cm} $$Using the tangent relation:
$$ \tan(2\theta) = \frac{\Delta y}{L} = \frac{48}{11 + R \cos \theta} \quad \text{…(2)} $$3. Solving the System
We have a system of two equations:
- $R = \frac{4}{\sin \theta}$
- $(11 + R \cos \theta) \tan(2\theta) = 48$
Substitute (1) into (2):
$$ \left( 11 + \frac{4}{\sin \theta} \cos \theta \right) \tan(2\theta) = 48 $$ $$ (11 + 4 \cot \theta) \left( \frac{2 \tan \theta}{1 – \tan^2 \theta} \right) = 48 $$This equation can be solved analytically, but checking for standard Pythagorean triples is often faster in such optics problems. Let’s test the 3-4-5 triangle where $R=5$ and $y=4$.
Test Case: $R = 5 \text{ cm}$
If $R=5$ and $y=4$, then:
$$ \sin \theta = \frac{4}{5} = 0.8 \implies \theta = 53.13^\circ $$ $$ \cos \theta = \frac{3}{5} = 0.6 $$Calculate the term $(11 + R \cos \theta)$:
$$ L = 11 + 5(0.6) = 11 + 3 = 14 \text{ cm} $$Calculate $\tan(2\theta)$:
$$ \tan(2\theta) = \frac{2 \tan \theta}{1 – \tan^2 \theta} $$ $$ \tan \theta = \frac{4}{3} \implies \tan(2\theta) = \frac{2(4/3)}{1 – (16/9)} = \frac{8/3}{-7/9} = -\frac{24}{7} $$The magnitude $|\tan(2\theta)| = \frac{24}{7} \approx 3.428$.
Now check the full equation:
$$ L \times |\tan(2\theta)| = 14 \times \frac{24}{7} = 2 \times 24 = 48 \text{ cm} $$The calculated vertical rise is exactly $48 \text{ cm}$. Adding this to the initial height $y=4$, we get $4 + 48 = 52 \text{ cm}$, which matches the given shadow radius.