Fermat's Principle, from which geometric optics can be derived, states that light travels by a path that minimizes the time of travel between the points. Consider a light beam that travels a horizontal distance \(D\) and a vertical distance \(h\), through two large flat slabs of material, with a vertical interface between the materials. One material has a thickness \(D / 2\) and index of refraction \(n_{1},\) and the second material has a thickness \(D / 2\) and index of refraction \(n_{2} .\) Determine the equation involving the indices of refraction and angles from horizontal that the light makes at the interface \(\left(\theta_{1}\right.\) and \(\theta_{2}\) ) which minimize the time for this travel.

Step by step solution

01

Compute distance the light travels in each slab

To compute the distance traveled by the light in each slab, we'll use the Pythagorean theorem. For slab 1, the horizontal distance is \(D/4\) and the vertical distance is \(h - x\) (where x is the vertical distance traveled by the light in the first slab). Thus, the distance traveled in slab 1 is: $$d_{1} = \sqrt{\left(\frac{D}{4}\right)^{2} + (h - x)^{2}}$$ Similarly, for slab 2, the horizontal distance is \(D/4\) and the vertical distance is \(x\). Therefore, the distance traveled in slab 2 is: $$d_{2} = \sqrt{\left(\frac{D}{4}\right)^{2} + x^{2}}$$

02

Compute the time it takes to travel those distances

To compute the time taken to travel through each slab, we need to divide the distance by the speed of light in the respective material. The speed of light in the material is given by \(v = \frac{c}{n}\), where \(c\) is the speed of light in vacuum and \(n\) is the index of refraction. The total time taken is the sum of the times taken in each slab. Let \(t_{1}\) and \(t_{2}\) be the times taken in slabs 1 and 2, respectively. Then: $$t_{1} = \frac{d_{1}}{v_{1}} = \frac{d_{1}}{\frac{c}{n_{1}}} = \frac{n_{1}d_{1}}{c}$$ $$t_{2} = \frac{d_{2}}{v_{2}} = \frac{d_{2}}{\frac{c}{n_{2}}} = \frac{n_{2}d_{2}}{c}$$ Total time: \(t = t_{1} + t_{2} = \frac{n_{1}d_{1} + n_{2}d_{2}}{c}\)

03

Apply Snell's Law to relate the angles and indices of refraction

Snell's law states that \(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\). Using the distances \(d_1\) and \(d_2\), we can relate the angles \(\theta_1\) and \(\theta_2\) to \(x\). The expressions for \(\sin\theta_1\) and \(\sin\theta_2\) in terms of \(x\) are given by: $$\sin\theta_{1} = \frac{h - x}{d_{1}}$$ $$\sin\theta_{2} = \frac{x}{d_{2}}$$

04

Minimize the time with respect to the angles

Using the expressions in Step 3, we have: $$n_{1} \frac{h-x}{d_{1}} = n_{2} \frac{x}{d_{2}}$$ This equation gives us the relation between the angles and the indices of refraction. Substituting the expressions for \(d_{1}\) and \(d_{2}\), we obtain: $$n_{1}\frac{h-x}{\sqrt{\left(\frac{D}{4}\right)^{2} + (h - x)^{2}}} = n_{2}\frac{x}{\sqrt{\left(\frac{D}{4}\right)^{2} + x^{2}}}$$ This is the desired equation involving the indices of refraction and angles that minimize the time for light to travel through the two slabs.

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