OPTICS BYU 31

Physics Solution Q31

Solution to Question 31

Objective: Find the radial variation of the refractive index $\mu(r)$ of a thin disc of thickness $d$ such that a parallel beam of light incident normally converges at a focal distance $f$.

Principle: Fermat’s Principle of Least Time (or constant optical path length). For the wavefronts to converge to a single focus $F$, the optical path length (OPL) for every ray from the incident plane wavefront to the focal point must be identical.

Optical Axis Disc (Thickness d) $\mu_0$ $\mu(r)$ $d$ $f$ $r$ F

Step 1: Optical Path Length Calculation

Let us compare the optical path length (OPL) of two rays:

  1. The Central Ray: Travels along the optical axis.
    • Path inside disc: Distance $d$, Refractive index $\mu_0$.
    • Path in air: Distance $f$.
    • $OPL_{center} = \mu_0 d + f$
  2. A Ray at distance $r$: Travels at a height $r$ from the axis.
    • Path inside disc: Distance $d$, Refractive index $\mu(r)$.
    • Path in air: Distance $\sqrt{f^2 + r^2}$ (hypotenuse from exit point to focus).
    • $OPL_{r} = \mu(r) d + \sqrt{f^2 + r^2}$

Step 2: Equating Optical Paths

For constructive interference at the focus (forming a sharp image), the optical path lengths must be equal:

$$ \mu_0 d + f = \mu(r) d + \sqrt{f^2 + r^2} $$

Rearranging to solve for $\mu(r)$:

$$ \mu(r) d = \mu_0 d + f – \sqrt{f^2 + r^2} $$ $$ \mu(r) = \mu_0 + \frac{f – \sqrt{f^2 + r^2}}{d} $$

Step 3: Approximation for Paraxial Rays ($f \gg r$)

Since the focal length is much larger than the radius of the disc ($f \gg r$), we can approximate the square root term using the binomial expansion $(1+x)^n \approx 1+nx$:

$$ \sqrt{f^2 + r^2} = f \left( 1 + \frac{r^2}{f^2} \right)^{1/2} \approx f \left( 1 + \frac{1}{2}\frac{r^2}{f^2} \right) = f + \frac{r^2}{2f} $$

Substitute this back into the equation for $\mu(r)$:

$$ \mu(r) = \mu_0 + \frac{f – (f + \frac{r^2}{2f})}{d} $$ $$ \mu(r) = \mu_0 + \frac{f – f – \frac{r^2}{2f}}{d} $$ $$ \mu(r) = \mu_0 – \frac{r^2}{2df} $$
Final Expression: $$ \mu(r) = \mu_0 – \frac{r^2}{2df} $$