As shown in Figure P22.49, a light ray is incident normal to...

As shown in Figure P22.49, a light ray is incident normal to one face of a $30^{\circ}-60^{\circ}-90^{\circ}$ block of dense flint glass (a prism) that is immersed in water. (a) Determine the exit angle $\theta_{4}$ of the ray, (b) A substance is dissolved in the water to increase the index of refraction. At what value of $n_{2}$ does total internal reflection cease at point $P_{f}$

Key Concepts

Snell's Law

Snell’s Law explains how light bends when it passes from one medium into another with a different index of refraction. It relates the sines of the angles of incidence and refraction to the indices of refraction of the two media, allowing calculation of the direction of the refracted ray at an interface.

Index of Refraction

The index of refraction is a measure of how much light slows down in a medium compared to its speed in a vacuum. It is a key parameter in determining how much light bends as it passes between different materials, and changes in its value can significantly affect the behavior of light at interfaces.

Prism Geometry

In optics, the geometry of a prism, including its angles, is crucial in determining how light is refracted through it. The specific internal angles of a prism dictate the path of light rays within the material and affect the overall deviation and exit angle of the ray after passing through the prism.

Total Internal Reflection

Total internal reflection occurs when a light ray traveling in a medium with a higher index of refraction strikes the boundary with a lower index medium at an angle greater than the critical angle, resulting in the light being completely reflected back into the original medium rather than being refracted.

Critical Angle

The critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90°. It marks the threshold beyond which total internal reflection occurs, and it is determined by the indices of refraction of the two media involved.

Transcript

0:00 Hi there.

00:01 So for this problem, as is shown in this figure, a light rate is incident normally to one phase of a 30 degrees, 60 degree, 90 degrees block of dense fling glass.

00:13 That means a prince that is immersed in water.

00:18 So for part a of this problem, we are asked about to determine the exit angle theta 4 of the ray, of the rate.

00:29 So we need to determine the determine that angle.

00:32 So from the geometry of the figure, as you can see, at the right, we can observe that teta 1 is equal to 60 degrees, because it's this same angle right here.

00:46 And so teta 1 is equal to 60 degrees.

00:52 Therefore, alpha, this angle alpha, is going to be equal to 90 degrees, this alpha is this angle right here.

01:04 This alpha, this is alpha.

01:07 So alpha is going to be 90 degrees minus theta 1.

01:12 So we're going to obtain that after we substitute that in here, alpha is going to be 30 degrees.

01:19 Now, we're going to have then that theta 2 plus 90 degrees.

01:33 Is plus alpha plus 30 degrees, that sum of all those angles will give us the angle of 180 degrees.

01:46 Then if we solve for the angle theta 2, we will obtain that theta 2 is equal to 180 degrees after we substitute those values, and minus 120 degrees minus alpha, and this will give us a value of 30 degrees.

02:11 So since the prince is immersing water, we know that the index of refraption 2 is going to be equal to 1 .333.

02:24 And with that set, oh sorry, the question in here should be the angle theta 3, because we don't have any angle teta 4.

02:38 So it should be angle teta 3...

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