PHYS 2002 Lecture : Physics Notes 26 3 26 8

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15 Mar 2019

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Consider a light ray in air striking the surface of water: Part of the ray is refracted, and part is reflected. The refracted ( r) and reflected ( r) angles will both change if we change the incident angle (law of reflection and snell"s law of refraction). Now consider what happens when a light ray emerges from a more dense medium into a less dense medium: water to air, for example. Part of the ray is reflected, and part is refracted. Notice that the refracted angle is greater than the incident angle since nair < nwater. Now let"s keep increasing the incident angle so that the refracted angle keeps increasing. Notice: there is a point when the refracted angle = 90o (red ray). This occurs when i = c, where c is known as the critical angle. Thus, when i = c, refracted = 900. Now, if i > c, then there is no refracted ray (purple ray)!

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

A ray of light in air is refracted as it goes into glass undergoes total internal reflection off of air at the bottom surface of the glass, and is then refracted again as it goes back up through the glass back out into the air. The top and bottom glass surfaces are parallel to each other. Which possible incident angles, will allow for this to occur?

olivecockroach604

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 2002 Lecture : Physics Notes 26 3 26 8

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