In Fig. 33-62, a light ray in air is incident at angle θ1 on a block of transparent plastic with

step 1 So here we have the ray diagram. We know that for part A, we need to use first Snell's Law of Re

In Fig. 33-62, a light ray in air is incident at angle θ1 on...

In Fig. $33-62,$ a light ray in air is incident at angle $\theta_{1}$ on a block of transparent plastic with an index of refraction of 1.56 . The dimensions indicated are $H=2.00 \mathrm{~cm}$ and $W=3.00 \mathrm{~cm} .$ The light passes through the block to one of its sides and there undergoes reflection (inside the block) and possibly refraction (out into the air). This is the point of first reflection. The reflected light then passes through the block to another of its sides - a point of second reflection. If $\theta_{1}=40^{\circ},$ on which side is the point of (a) first reflection and (b) second reflection? If there is refraction at the point of (c) first reflection and (d) second reflection, give the angle of refraction; if not, answer "none." If $\theta_{1}=70^{\circ},$ on which side is the point of (e) first reflection and (f) second reflection? If there is refraction at the point of ( $g$ ) first reflection and (h) second reflection, give the angle of refraction; if not, answer "none."

In Fig. $33-62,$ a light ray in air is incident at angle $\theta_{1}$ on a block of transparent plastic with an index of refraction of 1.56 . The dimensions indicated are $H=2.00 \mathrm{~cm}$ and $W=3.00 \mathrm{~cm} .$ The light passes through the block to one of its sides and there undergoes reflection (inside the block) and possibly refraction (out into the air). This is the point of first reflection. The reflected light then passes through the block to another of its sides - a point of second reflection. If $\theta_{1}=40^{\circ},$ on which side is the point of (a) first reflection and (b) second reflection? If there is refraction at the point of (c) first reflection and
(d) second reflection, give the angle of refraction; if not, answer "none." If $\theta_{1}=70^{\circ},$ on which side is the point of (e) first reflection and (f) second reflection? If there is refraction at the point of ( $g$ ) first reflection and (h) second reflection, give the angle of refraction; if not, answer "none."

Key Concepts

Snell's Law

Snell's Law is a fundamental principle in optics that relates the angles of incidence and refraction to the indices of refraction of two media. It is expressed mathematically as n? sin?? = n? sin??, where n? and n? are the indices of refraction for the first and second media, respectively, and ?? and ?? are the angles of incidence and refraction. This law explains how and why light bends when transitioning between different transparent materials.

Index of Refraction

The index of refraction is a measure of how much a medium slows down light relative to its speed in a vacuum. It plays a crucial role in determining the bending of light at interfaces, as characterized by Snell’s Law. Higher indices indicate that light travels slower in the medium, resulting in greater bending of the light ray when it enters or exits the material.

Total Internal Reflection

Total internal reflection occurs when light traveling in a medium with a higher index of refraction strikes an interface with a medium of lower index at an angle greater than the critical angle, resulting in no light being refracted out of the medium. Instead, the light is completely reflected back within the original medium. This phenomenon is a key principle in optical devices such as fiber optics.

Critical Angle

The critical angle is the specific angle of incidence in a medium above which total internal reflection occurs at the boundary with a less optically dense medium. It can be calculated using the formula sin?_c = n?/n?, where n? is the index of the initial medium and n? is the index of the second medium. Understanding the critical angle is essential for predicting whether light will refract or be reflected at an interface.

Law of Reflection

The Law of Reflection states that when a light ray reflects off a surface, the angle of incidence is equal to the angle of reflection. This rule applies to all reflective surfaces and is fundamental for tracing the path of light rays after reflection inside media such as glass or plastic.

Geometric Ray Tracing

Geometric ray tracing involves using geometric techniques to follow the paths of light rays as they reflect and refract at surfaces. This method is crucial for understanding the behavior of light in complex systems, such as how rays propagate through a transparent block with internal reflection and refraction, allowing one to predict the subsequent paths based on the system’s geometry.

Transcript

00:01 So here we have the ray diagram.

00:03 We know that for part a, we need to use first snell's law of refraction.

00:10 And so we can say that n sub 1 sign of theta sub 1 equals n sub 2 times sign of theta sub 2.

00:19 We can say that n sub 1 would be the refractive index of air, and n sub 2 would be the refractive index of the material, where theta sub 1 would be the angle at the ray entrance, and then theta sub 2 would be the angle at which the light ray strikes at the second surface.

00:39 We can say that for solving for theta sub 2, theta sub 2 being the angle at which the light ray strikes at the second surface, we can say that theta sub 2 would equal arc sign of n sub 1 sign of theta sub 1, divided by n sub 2.

01:02 We then know that theta sub 2 is equaling arc sign of 1 .0 sign of 40 degrees divided by n sub 2 of 1 .56.

01:18 This is equaling 24 .3 degrees.

01:28 Now we know that the diagonal angle is given by the figure here.

01:34 So we can say theta is going to be equal to the diagonal angle, arc tan of h divided by w.

01:42 We can say this would be arc tan of 2 centimeters divided by 3 centimeters, and this is equaling 33 .69 degrees.

01:53 We know that here 33 .69 degrees is, of course, less than theta sub 2.

02:03 Or rather my apologies, the value, here 33 .69 degrees is greater than theta sub 2.

02:20 So we know that here, light undergoes the reflection to the side 3.

02:41 So this would be your answer for part a.

02:44 Now, for part b, however, we know that here the light ray strikes at side 3.

02:52 And we're simply going to be used a trigonometric ratio at the upper right corner, and we can say tangent of theta is going to be equal to h over 3 .00 centimeters, or we can say, of course, h would be equal to three tangent of 24 .33 degrees.

03:12 This is giving us 1 .36 centimeters.

03:17 So we can say h prime would be equal to 2 centimeters minus 1 .3 .3 .3 centimeters, 36 centimeters giving us 0 .64 centimeters.

03:28 This would be the distance at which the ray strikes the side 3.

03:37 And so we can say that using the angle being symmetrical upon reflection, the ray strikes at the top of the surface at the point 1 .42 centimeters to the left of the corner of the square.

03:53 So we can say, upon reflection, ray strikes the top of surface at the point 1 .42 centimeters to the left corner of the square.

04:37 Given this, we know that the length 1 .42 centimeters is of course less than the three centimeters, and hence we can see.

04:46 Say that ray strikes at side two, which is the second reflection of the ray.

05:21 This would be your answer for part b.

05:23 For part c, now we know that here sides one and three are horizontal to one another and the angle of incidence is at side three.

05:35 Here the angle of incidence at the side three is the same as the angle reflection at side 1 so we can then hear you snell's law again 1 .56 times sign of 24 .3 degrees would be equal to n sub 1 sign of theta sub 1 and we can say then theta sub 1 would simply be equal to arc sign of 1 .56 sign of 24 .3 divided by 1 .23 divided by 1 .23 .3 degrees divided by 1 and we find that theta sub 1 is equaling 40 degrees.

06:21 This would be your answer for part b.

06:23 We know this would be the angle of refraction.

06:35 Again, theta sub 1 equals 40 degrees.

06:38 And finally, for part d, we know that, well, rather for part d, we know that the ray strikes surface 2.

06:53 This would be 90 degrees at an angle...

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