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Light in medium A undergoes a total internal reflection as it reaches the interface with medium B. Which of the following statements must be true (choose all that apply)? (a) nB < nA (b) nB > nA (c) All light rays that undergo a total internal reflection travel along the interface between the two materials. (d) Light traveling in the opposite direction, from B into A, cannot undergo a total internal reflection.

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Manoj Pandey
Manoj PandeyLv10
7 Mar 2021

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

Thepicture above shows a source of light in the lower medium (thepurple dot), with the light shining on the origin, which is markedby the position of the green dot. The horizontal axis is theinterface separating one medium from another. In this case, all ofthe light reflects back into the lower medium.
Choose all the correct statements about this situation from thelist below.
A.The lower medium (where the source is) has a larger index ofrefraction than the upper medium.
B.The lower medium has the same index of refraction as the uppermedium.
C.The lower medium has a smaller index of refraction than the uppermedium.
D.The angle of incidence in this case is greater than the criticalangle for total internal reflection to occur.
E.The angle of incidence in this case is smaller than the criticalangle for total internal reflection to occur.
F.If the light source was moved to the position of the red dot inthe upper medium, and the light was incident on the origin again,the light would definitely experience total internalreflection.
G.If the light source was moved to the position of the red dot inthe upper medium, and the light was incident on the origin again,some light would refract into the lower medium, traveling towardthe position of the purple dot.
H.If the light source was moved to the position of the red dot inthe upper medium, and the light was incident on the origin again,some light would refract into the lower medium, but the light wouldnot travel toward the position of the purple dot.

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 θr and the refracted angle θb. Since nb is greater than na, the refracted angle must be smaller than the incident angle θa; 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 θr= θa=60.0°.

To find the direction of the refracted ray, we use Snell’s law, Eq 33.4, with na=1.33, nb=1.42, and θa=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 θb< θa.

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.

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.

r

normal

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

The Law of Refraction –Snell’s Law

A ray of light passing from one substance (index ofrefraction

n1)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 θ1> 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 θc = 1/

n

1. (3)

Polarization byReflection – Brewster’s Angle

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

tan θ

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

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