Solution: Variable Refractive Index Sphere
1. Derivation of the Optical Invariant
To solve this problem, we must first derive the conservation law for a spherically symmetric medium, often called Bouguer’s Formula.
Consider the medium as a series of infinitesimally thin concentric spherical shells. Let a light ray pass from a shell of radius $r_1$ (refractive index $\mu_1$) into a shell of radius $r_2$ (refractive index $\mu_2$).
Figure 1: Geometry of a ray traversing concentric shells (A to B).
Step A: Snell’s Law at Point A
Applying Snell’s law at the interface at point A:
$$ \mu_1 \sin \phi_1 = \mu_2 \sin \phi’_1 \quad \text{— (i)} $$
Step B: Geometry of Triangle OAB
In the triangle $\Delta OAB$:
- Side $OA = r_1$
- Side $OB = r_2$
Step C: Combining Equations
Multiplying eq (ii) by $\mu_2$ and substituting $\mu_1 \sin \phi_1$ from eq (i):
$$ \mu_1 r_1 \sin \phi_1 = \mu_2 r_2 \sin \phi_2 $$
This implies the quantity is conserved along the path:
Optical Invariant: $$ \mu(r) \cdot r \cdot \sin \phi = \text{constant} $$
2. Applying the Invariant to the Problem
Given $\mu(r) \propto r \implies \mu(r) = k \cdot r$. Substituting this into the invariant:
$$ (kr) \cdot r \cdot \sin \phi = \text{Constant} \implies r^2 \sin \phi = C $$Figure 2: The trajectory of the ray passing through point P.
3. Calculation of Minimum Distance
Step A: Evaluate constant at point P
At point P, distance is $r = r_0$ and the angle between the ray and the radius vector is $\theta$ (as given in the diagram).
$$ C = r_0^2 \sin \theta $$
Step B: At Closest Approach ($r_{min}$)
At the minimum distance, the ray is perpendicular to the radius vector, so $\phi = 90^\circ$ and $\sin \phi = 1$.
$$ r_{min}^2 \sin(90^\circ) = C \implies r_{min}^2 = r_0^2 \sin \theta $$
$$ r_{min} = r_0 \sqrt{\sin \theta} $$
Step C: Numerical Solution
Values: $r_0 = 50\sqrt{2} \text{ cm}$, $\theta = 30^\circ$ ($\sin 30^\circ = 0.5$), Metal Sphere Radius $R = 10 \text{ cm}$.
$$ r_{min} = (50\sqrt{2}) \cdot \sqrt{0.5} = 50\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 50 \text{ cm} $$
The distance from the surface of the sphere is:
$$ \text{Distance} = r_{min} – R = 50 – 10 = 40 \text{ cm} $$