Solution to Question 18
1. Refraction at the First Surface
Consider a paraxial ray incident parallel to the axis at a height $r_b$. Parameters: $u_1 = \infty$, $R_1 = +r_s$, $\mu_1 = 1$, $\mu_2 = \mu = 2$. Using the spherical refraction formula: $$ \frac{\mu}{v_1} – \frac{1}{u_1} = \frac{\mu – 1}{R_1} $$ $$ \frac{2}{v_1} – 0 = \frac{2 – 1}{r_s} $$ $$ v_1 = 2 r_s $$ Given $r_s = 20.0$ cm, $v_1 = 40.0$ cm. Since the diameter of the sphere is $2r_s = 40.0$ cm, the rays converge exactly at the back pole of the sphere.
2. Refraction at the Second Surface
The rays attempt to converge at the pole of the second surface. This acts as a virtual object or simply the point of emergence. Let’s calculate the angle of incidence at the back surface. The ray travels a horizontal distance $2r_s$ while dropping a height $r_b$. The angle of the ray inside the sphere with the axis is: $$ \theta \approx \tan \theta = \frac{r_b}{v_1} = \frac{r_b}{2r_s} $$ At the back pole, the normal lies along the optical axis. Thus, the angle of incidence $i = \theta$. Using Snell’s Law for emergence into air ($\mu \to 1$): $$ \mu \sin i = 1 \cdot \sin e $$ Using small angle approximation: $$ e = \mu i = \mu \left( \frac{r_b}{2r_s} \right) $$ With $\mu = 2$: $$ e = 2 \cdot \frac{r_b}{2r_s} = \frac{r_b}{r_s} $$
3. Calculating Spot Size on Screen
The rays diverge from the back pole of the sphere. Distance from sphere centre to screen is $l = 100$ cm. Distance from back pole to screen is $D = l – r_s = 100 – 20 = 80$ cm. The radius of the spot $R$ is determined by the divergence angle $e$: $$ R = D \tan e \approx D \cdot e $$ Substituting $e = r_b / r_s$: $$ R = (l – r_s) \frac{r_b}{r_s} $$
4. Numerical Substitution
$$ l = 100 \text{ cm}, r_s = 20 \text{ cm}, r_b = 0.5 \text{ cm} $$ $$ R = (100 – 20) \times \frac{0.5}{20} $$ $$ R = 80 \times 0.025 = 2.0 \text{ cm} $$