Question

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

Short answer

Answer: The equation relating the indices of refraction and the angles, which minimize the time of travel through two slabs of material with different refractive indices, is given by:\(\frac{\sin\theta_{1}}{\cos^2\theta_{1}}=n_{2}\frac{\sin\theta_{2}}{\cos^2\theta_{2}}\)

Step-by-step solution

Draw a diagram

Draw a diagram of the problem, with the two slabs of material, and the light beam entering at angle \(\theta_{1}\) and refracting through the interface and into the second slab at angle \(\theta_{2}\). Be sure to label the angles, thicknesses, and refractive indices.

Write the equation for time, distance, and speed

Since we are to find the path that minimizes the time of travel, we have to write an equation for the time. Recall that time = distance/speed and that the speed of light is given by \(v=c/n\), where \(c\) is the speed of light and \(n\) is the index of refraction. In each of the two slabs, we can write expressions for distance in terms of the given horizontal and vertical distances, \(D\) and \(h\), as well as the angles \(\theta_{1}\) and \(\theta_{2}\). Let's consider the distances traveled by light: In the first slab: \(x=\frac{D}{2}\tan\theta_{1}\) and the total distance traveled is \(L_{1} = \sqrt{\left(\frac{D}{2}\right)^2 + h^2}\) In the second slab: \(y=\frac{D}{2}\tan\theta_{2}\) and the total distance traveled is \(L_{2} = \sqrt{\left(\frac{D}{2}\right)^2 + y^2}\)

Find the time of travel in each slab

Now that we have the expressions for the distance, we can calculate the time of travel in each slab and add them to get the total time. Time in the first slab: \(T_{1}=\frac{L_{1}}{c/n_{1}}=n_{1}\frac{L_{1}}{c}\) Time in the second slab: \(T_{2}=\frac{L_{2}}{c/n_{2}}=n_{2}\frac{L_{2}}{c}\) The total time of travel is the sum of \(T_{1}\) and \(T_{2}\), which is: \(T=T_{1} + T_{2} = n_{1}\frac{L_{1}}{c} + n_{2}\frac{L_{2}}{c}\) We want to find the minimum time, so we will minimize the relation: \(\frac{dT}{d\theta_{1}}=0\)

Use Snell's Law

Snell's Law states that the product of the refractive index \(n\) and the sine of the angle of incidence \(\theta\) is constant when crossing an interface between materials. This can be written as: \(n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}\) We can use this to find an equation that relates the indices of refraction and the angles, which minimize the time of travel.

Substitute and find the relationship

Substitute the relationship from Snell's Law (Step 4) into the equation we found in Step 3. After simplifying and differentiating with respect to \(\theta_{1}\), we get: \(\sin\theta_{1}(\frac{1}{\sqrt{1+\tan^2\theta_{1}}})=n_{2}\sin\theta_{2}(\frac{1}{\sqrt{1+\tan^2\theta_{2}}})\) Further simplification gives us the desired equation: \(\frac{\sin\theta_{1}}{\cos^2\theta_{1}}=n_{2}\frac{\sin\theta_{2}}{\cos^2\theta_{2}}\) So, this is the equation relating the indices of refraction and the angles \(\theta_{1}\) and \(\theta_{2}\) which minimize the time of travel.