In Fig. 33-78, where n1=1.70 n2=1.50, and n3=1.30, light...

In Fig. $33-78,$ where $n_{1}=1.70$ $n_{2}=1.50,$ and $n_{3}=1.30,$ light refracts from material 1 into material 2 . If it is incident at point $A$ at the critical angle for the interface between materials 2 and 3 , what are (a) the angle of refraction at point $B$ and (b) the initial angle $\theta ?$ If, instead, light is incident at $B$ at the critical angle for the interface between materials 2 and $3,$ what are $(\mathrm{c})$ the angle of refraction at point $A$ and (d) the initial angle $\theta ?$ If, instead of all that, light is incident at point $A$ at Brewster's angle for the interface between materials 2 and $3,$ what are (e) the angle of refraction at point $B$ and (f) the initial angle $\theta ?$

In Fig. $33-78,$ where $n_{1}=1.70$ $n_{2}=1.50,$ and $n_{3}=1.30,$ light refracts from material 1 into material 2 . If it is incident at point $A$ at the critical angle for the interface between materials 2 and 3 , what are (a) the angle of refraction at point $B$ and
(b) the initial angle $\theta ?$ If, instead, light is incident at $B$ at the critical angle for the interface between materials 2 and $3,$ what are $(\mathrm{c})$ the angle of refraction at point $A$ and (d) the initial angle $\theta ?$ If, instead of all that, light is incident at point $A$ at Brewster's angle for the interface between materials 2 and $3,$ what are (e) the angle of refraction at point $B$ and (f) the initial angle $\theta ?$

Key Concepts

Snell's Law

Snell's Law is the fundamental principle that governs how light refracts when it passes from one medium into another with different refractive indices. It formalizes the relationship between the angles of incidence and refraction and the refractive indices of the two media, allowing us to predict the change in direction of a light ray as it transitions between materials.

Critical Angle and Total Internal Reflection

The critical angle is the minimum angle of incidence at which light traveling from a medium with a higher refractive index to one with a lower refractive index undergoes total internal reflection rather than being refracted. This concept is crucial for understanding limits in light transmission at interfaces and is applied in scenarios where light must be confined within a medium, as seen in fiber-optic technology.

Brewster's Angle

Brewster's angle is the angle of incidence at which light with a specific polarization is completely transmitted through the interface between two media without any reflection. At this angle, the reflected and refracted rays are perpendicular to each other, and this phenomenon is utilized in optical devices such as polarizers to minimize glare and enhance image clarity.

Transcript

00:01 For this problem on the topic of electromagnetic waves, in the figure we are given the refractive indices for the three materials, n1, n2, and n3, which are 1 .7, 1 .5, and 1 .3 respectively, we are told that light refracts from material 1 into material 2.

00:17 And if it is incident at point a at the critical angle for the interface between materials 2 and 3, we want to find the angle of refraction at point b, the initial angle theta, and then we are told that if instead light is incident at b at the critical angle between the materials 2 and 3, we want to find the angle of refraction at a and the initial angle theta.

00:43 Lastly, if instead the light is incident at a at brewster's angle for the interface between 2 and 3, we want to again find the angle of refraction now at point b and the initial angle theta.

00:56 Now the angle of incidence theta b1 at b is the complement of the critical angle at a and so it's sign sign of theta b1 is equal to the cosine of the critical angle theta c which is equal to the square root of 1 minus n3 over n2 all squared.

01:29 And so the angle of refraction theta b2 at b becomes the arc sign of n2 over n3 times the square root of 1 minus n3 over n2 all squared.

01:55 And so this simplifies to the arc sign of the square root of n2 over n2 over n3.

02:07 All squared minus 1, putting in our values for the refractive indices, we get this to be 35 .1 degrees.

02:21 Now for part b, we know that n1 sine, side theta, in this case, is equal to n2 sign of the critical angle theta c, which is n2 into n3 divided by n2.

02:46 And so we get theta equal to the arc sign of n3 over n1.

02:59 Again, if you put these values in, we get this to be 49 .9 degrees.

03:09 Now for part c, the angle of incidence theta a1 at a is the complement of the critical angle at b...

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