Physics 1C Lecture 25B. Mirages Mirages occur because of total internal reflection (TIR). Hot air is less dense than cool air. The less dense hot air.

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Presentation transcript:

Physics 1C Lecture 25B

Mirages Mirages occur because of total internal reflection (TIR). Hot air is less dense than cool air. The less dense hot air has a small index of refraction. The more dense cool air has a higher index of refraction. TIR happens for light in cool air (slow) off of hot air (fast).

Fata Morgana

Huygens’s Principle Christiaan Huygens ( ) introduced a geometric construction that helped to understand many practical aspects of propagation of light. In this construction, all points of a wave front are taken as point sources for the production of secondary waves, called wavelets. New position of the wave front after time  t has elapsed is the surface tangent to the wavelets.

Huygens’s Principle We can use Huygens’s principle to derive the laws of reflection and refraction. AB is a wave front of incident light. Point A sends out a wavelet toward point D. Point B sends out a wavelet toward point C. Both rays move with the same speed: AD = BC = c  t Let’s isolate congruent triangles ABC and ADC

Huygens’s Principle ABC and ADC are congruent because they share hypotenuse AC and because AD = BC. We have: cos  = BC / AC and cos  ´ = AD / AC. Therefore, cos  = cos  ’ and  =  ’ This gives θ 1 = θ 1 ’ which is the law of reflection

Prisms Example An incident ray in air is headed straight towards an equilateral plastic prism (n = 1.50). The ray is parallel to the bottom of the prism. Use Snell’s Law to find the angle (with respect to the normal) that the light ray exits the prism on the right. Answer The diagram is given but it is up to you to draw the normal(s) and path of the ray.

Prisms Answer Start with: Thus, the incident angle is 30 o from air to plastic. Using Snell’s Law we find: 60 o 30 o θ2θ2

Prisms Answer Now, we need to look at the second boundary. We can now examine the small top triangle created by the ray in the prism. The bottom left angle on this triangle will be: This means that the bottom right angle of the triangle will be: 60 o o o θ2θ2

Prisms Answer Is the o, the angle we will use in Snell’s Law? No, it is not with respect to the normal. We draw the normal and find: At the second boundary, will this ray be refracted or totally internally reflected? Check by calculating the critical angle. 60 o o θ3θ3 refracted, barely

Prisms Answer Now we need to calculate the outgoing ray with Snell’s Law again: θ3θ3 θ4θ4 This is the outgoing angle with respect to the normal

Prisms Answer Overall, this incident ray will be pushed downward compared to its original direction. Don’t forget your geometry when dealing with prisms and Snell’s Law. θ 1 +θ 2 = 180 o θ 1 +θ 2 = 90 o θ 1 +θ 2 +θ 3 =180 o

Concept Question In the prism example we just performed, a light ray was deflected downward when it moved through a prism when the pointy side was up. If you inverted the prism (so the pointy side was down) how would the direction of the incident ray change after passing completely through the front and back sides of the prism? A) The light ray would still be deflected downward. B) The light ray would now be deflected upward. C) The light ray would not be deflected (it would move in the same direction that it had originally).

Dispersion The index of refraction in anything except a vacuum depends on the wavelength of the light. The dependence of n on λ is called dispersion. Snell’s Law indicates that the angle of refraction made when light enters a material depends on the wavelength of the light. The index of refraction for a material usually decreases with increasing wavelength.

Dispersion This means that violet light refracts more than red light when passing from air into a material. The amount the ray is bent away from its original direction is called the angle of deviation, δ. Since all the colors have different angles of deviation, they will spread out into a spectrum. Violet deviates the most. Red deviates the least.

Rainbows Rainbows appear when a ray of light strikes a drop of water in the atmosphere. The incident white light ray will undergo both reflection and refraction. At the first boundary (air-water), refraction will occur (with red deviating the least). At the second boundary (water-air), reflection will occur. It refracts again as it hits a third boundary (water-air).

Rainbows The angle between the incident white light and the violet ray is 40 o. The angle between the incident white light and the red ray is 42 o. All of the other visible colors will lie in between violet and red. When you see a rainbow in the sky you are actually seeing this happen in many, many raindrops.

Rainbows For a raindrop very high in the sky, red will be bent more and will be directed toward the observer. A raindrop lower in the sky would direct violet toward the observer.

For Next Time (FNT) Start reading Chapter 26 Start working on the homework for Chapter 25