Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A ray of light travels from point \(P_{1}\) in a medium of refractive index \(n_{1}\) to \(P_{2}\) in a medium of index \(n_{2},\) by way of the point \(Q\) on the plane interface between the two media, as in Figure \(6.9 .\) Show that Fermat's principle implies that, on the actual path followed, \(Q\) lies in the same vertical plane as \(P_{1}\) and \(P_{2}\) and obeys Snell's law, that \(n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2} .\) [Hints: Let the interface be the \(x z\) plane, and let \(P_{1}\) lie on the \(y\) axis at \(\left(0, h_{1}, 0\right)\) and \(P_{2}\) in the \(x, y\) plane at \(\left(x_{2},-h_{2}, 0\right) .\) Finally let \(Q=(x, 0, z)\) Calculate the time for the light to traverse the path \(P_{1} Q P_{2}\) and show that it is minimum when \(Q\) has \(z=0\) and satisfies Snell's law.]

Short Answer

Expert verified
Q lies on the plane interface, obeying Snell's law: \(n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}\).

Step by step solution

01

Setting up the geometry

The ray of light travels from \(P_{1}\) at \((0, h_{1}, 0)\) to \(Q\) at \((x, 0, z)\), and then to \(P_{2}\) at \((x_{2}, -h_{2}, 0)\). The plane interface is the \(xz\)-plane (\(y = 0\)). We assume the path from \(P_{1}\) to \(Q\) is in medium 1 with refractive index \(n_{1}\), and from \(Q\) to \(P_{2}\) is in medium 2 with refractive index \(n_{2}\).
02

Calculating the path lengths

The distance from \(P_{1}\) to \(Q\) is \(L_{1} = \sqrt{x^2 + h_{1}^2 + z^2}\) and the distance from \(Q\) to \(P_{2}\) is \(L_{2} = \sqrt{(x_{2} - x)^2 + h_{2}^2 + z^2}\).
03

Expressing time of travel

The time to travel from \(P_{1}\) to \(P_{2}\) via \(Q\) is \(T = \frac{n_{1}L_{1}}{c} + \frac{n_{2}L_{2}}{c}\), where \(c\) is the speed of light in a vacuum. Simplified, it becomes \(T = n_{1}\sqrt{x^2 + h_{1}^2 + z^2} + n_{2}\sqrt{(x_{2} - x)^2 + h_{2}^2 + z^2} \).
04

Applying Fermat's principle

According to Fermat's principle, light follows the path which takes the least time. This means we need to minimize \(T\) with respect to \(x\) and \(z\).
05

Setting condition for minimal time with respect to z

For time \(T\) to be minimal, its partial derivative with respect to \(z\) must be zero: \(\frac{\partial T}{\partial z} = 0\). Solving, due to symmetry and reflection principles in optics, leads to \(z = 0\), meaning \(Q\) lies in the plane interface (\(xz\)-plane).
06

Setting condition for minimal time with respect to x and deriving Snell's law

The partial derivative \(\frac{\partial T}{\partial x} = 0\) condition leads to Snell's law. Differentiating \(T\) and setting the derivatives to zero, we find that \[n_{1} \frac{x}{\sqrt{x^2 + h_{1}^2}} = n_{2} \frac{x - x_{2}}{\sqrt{(x_{2} - x)^2 + h_{2}^2}}\].Dividing both sides by the path-length components, it becomes \[n_{1}\frac{\sin \theta_{1}}{\cos \theta_{1}} = n_{2}\frac{\sin \theta_{2}}{\cos \theta_{2}}\], thus \[n_{1}\sin \theta_{1} = n_{2}\sin \theta_{2}\].
07

Conclusion

From minimizing the time of travel, we've shown \(Q\) lies on the \(xz\)-plane, proving Snell's law: \(n_{1} \sin \theta_{1}= n_{2} \sin \theta_{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Snell's Law
Snell's Law is a fundamental principle in optics that describes how light rays bend when they pass from one medium to another. This bending occurs because light travels at different speeds in different materials, which is determined by their refractive indices. Snell's Law is mathematically expressed as:

\[n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\]
Where:\
    \
  • \(n_{1}\) and \(n_{2}\) are the refractive indices of the two media\
  • \(\theta_{1}\) and \(\theta_{2}\) are the angles of incidence and refraction, respectively.
This law is crucial in understanding how light behaves when transitioning between air and water, glass, or any other transparent materials. It shows not just the relationship between the angles and refractive indices but also plays a critical role in explaining phenomena such as refraction, which is why lenses can focus light and why objects under water appear distorted.
Refractive Index
The refractive index, often represented by the symbol \(n\), is a dimensionless number that describes how fast light travels through a material. It is defined as the ratio of the speed of light in vacuum to its speed in the specified medium:

\[n = \frac{c}{v}\]
Where:
  • \(c\) is the speed of light in vacuum (approximately \(3 \times 10^8\) m/s)
  • \(v\) is the speed of light in the medium
A medium with a higher refractive index will slow down the light more than a medium with a lower one, resulting in the bending of the light path as per Snell's Law. This concept is essential in designing optical instruments, such as glasses and cameras because it allows us to predict and control how light will behave as it passes through various lenses and prisms. Understanding refractive indices explains not only the bending of light but also phenomena like magnification and the dispersion of light into its constituent colors.
Light Path Optimization
Light path optimization refers to the idea that light takes the path that requires the least time, also known as Fermat's Principle. This principle is the foundation of why Snell's Law holds true and is evident in the way light refracts.

The idea is that light will always choose the quickest available route between two points, which isn't always a straight line when multiple media are involved. Instead, it may deviate to minimize travel time across different materials with different refractive indices. This concept is crucial, as it elucidates why we see the bending of light at boundaries, thus forming the core of optical phenomena.
When dealing with lenses or fiber optics, grasping this optimization helps in designing systems that efficiently manage light paths. For instance:
  • Lenses are crafted to ensure light converges at a focus point, optimizing clarity and precision.
  • Fiber optic cables are engineered with specific refractive indices to keep light traveling within the cable's core, enhancing data transmission rates.
Understanding light path optimization gives us the ability to use light energy more efficiently, erring on the principles of time minimization in complicated light routing scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a right circular cylinder of radius \(R\) centered on the \(z\) axis. Find the equation giving \(\phi\) as a function of \(z\) for the geodesic (shortest path) on the cylinder between two points with cylindrical polar coordinates \(\left(R, \phi_{1}, z_{1}\right)\) and \(\left(R, \phi_{2}, z_{2}\right) .\) Describe the geodesic. Is it unique? By imagining the surface of the cylinder unwrapped and laid out flat, explain why the geodesic has the form it does.

Find the equation of the path joining the origin \(O\) to the point \(P(1,1)\) in the \(x y\) plane that makes the integral \(\int_{o}^{P}\left(y^{2}+y y^{\prime}+y^{2}\right) d x\) stationary.

An aircraft whose airspeed is \(v_{\mathrm{o}}\) has to fly from town \(O\) (at the origin) to town \(P\), which is a distance \(D\) due east. There is a steady gentle wind shear, such that \(\mathbf{v}_{\text {wind }}=V y \hat{\mathbf{x}},\) where \(x\) and \(y\) are measured east and north respectively. Find the path, \(y=y(x),\) which the plane should follow to minimize its flight time, as follows: (a) Find the plane's ground speed in terms of \(v_{o}, V, \phi\) (the angle by which the plane heads to the north of east), and the plane's position. (b) Write down the time of flight as an integral of the form \(\int_{0}^{D} f d x .\) Show that if we assume that \(y^{\prime}\) and \(\phi\) both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand \(f\) takes the approximate form \(f=\left(1+\frac{1}{2} y^{2}\right) /(1+k y)\) (times an uninteresting constant) where \(k=V / v_{\mathrm{o}}\). (c) Write down the Euler-Lagrange equation that determines the best path. To solve it, make the intelligent guess that \(y(x)=\lambda x(D-x),\) which clearly passes through the two towns. Show that it satisfies the EulerLagrange equation, provided \(\lambda=(\sqrt{4+2 k^{2} D^{2}}-2) /\left(k D^{2}\right) .\) How far north does this path take the plane, if \(D=2000\) miles, \(v_{\mathrm{o}}=500 \mathrm{mph},\) and the wind shear is \(V=0.5 \mathrm{mph} / \mathrm{mi} ?\) How much time does the plane save by following this path? [You'll probably want to use a computer to do this integral.]

Consider a ray of light traveling in a vacuum from point \(P_{1}\) to \(P_{2}\) by way of the point \(Q\) on a plane mirror, as in Figure \(6.8 .\) Show that Fermat's principle implies that, on the actual path followed, \(Q\) lies in the same vertical plane as \(P_{1}\) and \(P_{2}\) and obeys the law of reflection, that \(\theta_{1}=\theta_{2} .[\) Hints: Let the mirror lie in the \(x z\) plane, and let \(P_{1}\) lie on the \(y\) axis at \(\left(0, y_{1}, 0\right)\) and \(P_{2}\) in the \(x y\) plane at \(\left(x_{2}, y_{2}, 0\right)\) Finally let \(Q=(x, 0, z)\). Calculate the time for the light to traverse the path \(P_{1} Q P_{2}\) and show that it is minimum when \(Q\) has \(z=0\) and satisfies the law of reflection.]

Consider a medium in which the refractive index \(n\) is inversely proportional to \(r^{2}\); that is, \(n=a / r^{2},\) where \(r\) is the distance from the origin. Use Fermat's principle, that the integral (6.3) is stationary, to find the path of a ray of light travelling in a plane containing the origin. [Hint: Use twodimensional polar coordinates and write the path as \(\phi=\phi(r) .\) The Fermat integral should have the form \(\int f\left(\phi, \phi^{\prime}, r\right) d r,\) where \(f\left(\phi, \phi^{\prime}, r\right)\) is actually independent of \(\phi .\) The Euler-Lagrange equation therefore reduces to \(\partial f / \partial \phi^{\prime}=\) const. You can solve this for \(\phi^{\prime}\) and then integrate to give \(\phi\) as a function of \(r .\) Rewrite this to give \(r\) as a function of \(\phi\) and show that the resulting path is a circle through the origin. Discuss the progress of the light around the circle.]

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free