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 $\mathrm{A}$ 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 (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 $\underline{\theta}$ ?

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 $\mathrm{A}$ 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
(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 $\underline{\theta}$ ?

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, we are shown a figure with three refractive indices, n1, n2, and n3, which are 1 .7, 1 .5, and 1 .3, respectively, in which light is refracting from material 1 into material 2.

00:15 It is incident at a at the critical angle of the interface between materials 2 and 3, and we want to find the angle of refraction at b, the initial angle theta, and then if instead the light is incident at b, at the critical angle for the interface between 2 and 3, we want to find the angle of refraction at point a and the initial angle theta.

00:37 And lastly, if instead of all of that, the light is incident at point a at brewster's angle for the interface between materials 2 and 3, we want to find the angle of refraction at b and the initial angle theta.

00:50 Now, firstly for a, since the angle of incidence, theta b1 at b is the complement of the critical angle at a, its sign is as follows.

01:04 Sign of theta b1 is equal to cosine of the critical angle theta c, which is equal to the square root of 1 minus n3 over n2 all squared.

01:24 So the angle of refraction theta b2 at b then becomes the arc sign of n2 over n3 times the square root of 1 minus n3 over n2 all squared.

01:52 Which is the arc sign of the square root of n2 over n3 all squared minus 1.

02:12 And putting in our values, we get this angle to be 35 .1 degrees.

02:24 Now from part b, we can use n1 sine theta equal to n2.

02:34 Sign of the critical angle theta c, and this is equal to n2 times n3 over n2, which gives us the angle theta that we require to be the arc sign of n3 over n1, which gives us the angle 49 .9 degrees.

03:01 For part c, the angle of incidence theta a1 at a is the complement of the critical angle at b, and so it's sign, sine of theta a1, is the cosine of the critical angle theta c, which is the square root of 1 minus n3 over into all squared.

03:34 So the angle of reflection theta a 2 at a can be therefore found as the arc sign of n2 over n3 multiplied by the square root of 1 minus n3 over n2 all squared, which we can write as the arc sign of the square root of n2 over n3 all squared minus 1, which gives us the angle of refraction at a to be 35 .1 degrees.

04:29 For part d, we can apply snows low again...

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