1. Analyzing the Gradient Index

Light passes through a rectangular slab of length $l = 20 \text{ cm}$ and thickness $d = 4.0 \text{ cm}$. The refractive index $\mu$ increases linearly from bottom to top.

The total change in refractive index is $\delta \mu = 2.0 \times 10^{-4}$ over the thickness $d$.

The gradient of the refractive index is:

$$ \frac{d\mu}{dy} = \frac{\delta \mu}{d} = \frac{2.0 \times 10^{-4}}{4.0} = 0.5 \times 10^{-4} \text{ cm}^{-1} $$

2. Optical Path Length and Wavefront Tilt

As the parallel beam travels a distance $l$ through the slab, the optical path length (OPL) varies with height $y$:

$$ \text{OPL}(y) = \mu(y) \cdot l $$

The difference in OPL between the top and bottom of the beam is:

$$ \Delta \text{OPL} = (\Delta \mu) \cdot l = (2.0 \times 10^{-4}) \times 20 \text{ cm} = 4.0 \times 10^{-3} \text{ cm} $$

This variation in path length causes the planar wavefront to tilt. The angle of tilt $\theta$ (in radians) is the path difference divided by the beam width (thickness $d$):

$$ \theta = \frac{\Delta \text{OPL}}{d} = \frac{4.0 \times 10^{-3}}{4.0} = 1.0 \times 10^{-3} \text{ rad} $$

Alternatively, using the gradient directly: $\theta = \frac{d\mu}{dy} l = (0.5 \times 10^{-4})(20) = 10^{-3} \text{ rad}$.

3. Position of the Light Spot

The lens of focal length $f = 20 \text{ cm}$ focuses this collimated but tilted beam to a point in its focal plane. The shift $\Delta y$ from the central axis is:

$$ \Delta y = f \cdot \theta $$

Substituting the values:

$$ \Delta y = 20 \text{ cm} \times 1.0 \times 10^{-3} = 20 \times 10^{-3} \text{ cm} = 2.0 \times 10^{-2} \text{ cm} $$

Since the refractive index is higher at the top, the speed of light is lower at the top. The wavefronts are retarded at the top, causing the beam to tilt upwards towards the region of higher refractive index.