PHYS 142 Lecture Notes - Lecture 35: Specular Reflection, Diffuse Reflection

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7 Apr 2015

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Phys 142: electromagnetism and optics - lecture 35 : ray optics part 1. A light ray is defined as a line in the direction along which light energy is flowing. Any narrow beam of light, such as a laser beam, is actually a bundle of many parallel light rays. We can think of a single light ray as the limiting case of a laser beam whose diameter approaches zero. Light travels through a transparent material in straight lines called light rays. The speed of light is v = c/n, where n is the index of refraction of the material. Light rays do not interact with each other. Two rays can cross without either being affected in any way. Light interacts with matter in four different ways: At an interface between two materials, light can be either reflected or refracted. o: within a material, light can be either scattered or absorbed.

Related Questions

In Example 33.11, the water-glass interface is horizontal. If instead this interface were tilted 15.0° above the horizontal, with the right side higher than the left side, what would be the angle from the vertical of the ray in the glass? (The ray in the water still makes an angle of 60.0° with the vertical).

EXAMPLE 33.11: Reflection and refraction

In Fig 33.11, material a is water and material b is a glass with an index of refraction 1.52. If the incident ray makes an angle of 60.0° with the normal, find the directions of the reflected and refracted rays.

SOLUTION

IDENTIFY: This is a problem in geometric optics. We are given the incident angle and the index of refraction of each material, and we need to find the reflected and refracted angles.

SET-UP: Figure 33.11 shows the rays and angles for this situation. The target variables are the reflected angle θ_r and the refracted angle θ_b. Since n_b is greater than n_a, the refracted angle must be smaller than the incident angle θ_a; this is shown in the figure.

EXECUTE: According to Eq 33.2, the angle the reflected ray makes with the normal is the same as that of the incident ray so θ_r= θ_a=60.0°.

To find the direction of the refracted ray, we use Snell’s law, Eq 33.4, with n_a=1.33, n_b=1.42, and θ_a=60.0°. We find:

EVALUATE: The second material has a larger refractive index than the first, just like the situation shown in Fig 33.8a. Hence, the refracted ray is bent toward the normal as the wave slows down upon entering the second material and θ_b< θ_a.

Figure 33.11. Reflection and refraction of light passing from water to glass

Figure 33.8. Reflection and refraction in three cases. (a) Material b has a larger index of refraction than material a. (b) (a) Material b has a smaller index of refraction than material a. (c) The incident light ray is normal to the interface between the materials.

burgundysnake577

You will be utiizing a basicoptic system.

The index of refraction of theplastic cylindrical lens depends on wavelength. Choose an anglewhich gives a good color separation and measure the index ofrefraction for blue and red light.

Question:How big is the difference in indices of refraction? How good is theapproximation that n is constant for thismaterial?

If this question doesnt make sense..then just answer ittheoretically speaking. That is how would we go about checkingwhether or not it conforms with the n as the constant for thematerial... Thanks!

Background info:

ThePhysics

The Law ofReflection

The law of reflection states that for a plane mirror, the angle ofincidence is equal to the angle of reflection, and that the tworays lie in a plane.

normal

. A light ray perpendicular to the surface has an angle ofincidence 0Â°.

The Law of Refraction âSnellâs Law

A ray of light passing from one substance (index ofrefraction

n_{1)into another (index of refraction n}

2) across a plane boundary will be refracted. The angle the raymakes in one medium will be related to the angle in the secondmedium by the equation below, and that all rays lie in aplane.

(2)

Total InternalReflection

A ray incident on a medium with lower index of refraction willbe

totally internally reflected if sin Î¸1> n2/n1.There will be no transmitted beam for this incident angle. Ameasurement of the critical angle for total internalreflection can be used to determine the index of refraction of the substance.If n

2 = 1 (air), then the critical angle is:

sin Î¸c = 1/

1. (3)

Polarization byReflection â Brewsterâs Angle

If the angle between the reflected and refracted rays is 90Â°,then

tan Î¸

B = n2/n

1. (4)

At this angle, only light with polarization out of the plane ofreflection and refraction will be reflected. The reflected lightwill be completely polarized.

Figure3.

Refractionof Light Rays

Figure2.

Incidentand Reflected Rays

(1)

Always measure the angles with respect to the surface

PHYS 142 Lecture Notes - Lecture 35: Specular Reflection, Diffuse Reflection

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