Name: Sources of Electromagnetic Waves
Uploaded: 2019-07-19
Duration: 5 min 12 s
Description: Can you name some sources of electromagnetic waves by looking around you? You will learn about the different sources of electromagnetic waves

Question

Suppose the prism of the figure (a) has apex angle $ϕ = 60.0 °$ and index of refraction $n = 1.56 .$ (a) What is the smallest angle of incidence $θ$ for which a ray can enter the left face of the prism and exit the right face? (b) What angle of incidence $θ$ is required for the ray to exit the prism with an identical angle $θ$ for its refraction, as it does in the figure (b)?

(a)

(b)

Accepted Answer

(a) We refer to the entry point for the original incident ray as point $A$ (which we take to be on the left side of the prism as in the figure), the prism vertex as point $B,$ and the point where the interior ray strikes the right surface of the prism as point $C$ . The angle between line $A B$ and the interior ray is $β$ (the complement of the angle of refraction at the first surface) and the angle between the line $B C$ and the interior ray is $θ$ (the complement of its angle of incidence when it strikes the second surface). When the incident ray is at the minimum angle for which light is able to exit the prism, the light exits along the second face. That is, the angle of refraction at the second face is $θ$ and the angle of incidence there for the interior ray is the critical angle for total internal reflection. Let $θ_{1}$ be the angle of incidence for the original incident ray and $θ_{2}$ be the angle of refraction at the first face and let $θ_{3}$ be the angle of incidence at the second face. The law of refraction, applied to point $C$ yields $n \sin θ_{3} = 1,$ so

$\sin θ_{3} = \frac{1}{n} = \frac{1}{1.60} = 0.625 \Rightarrow θ_{3} = 38.68 °$

The interior angles of the triangle $A B C$ must sum to $180 °,$ so $α + β = 120 ° .$ Now, $α = 90 ° - θ_{3} = 51.32 °,$ so $β = 120 ° - 51.32 ° = 69.68 ° .$ Thus, $θ_{2} = 90 ° - β = 21.32 ° .$ The law of refraction, applied to point $A,$ yields

$\sin θ_{1} = n \sin θ_{2} = 1.60 \sin 21.32 ° = 0.5817$

Thus, $θ_{1} = 35.6 °$

(b) We apply the law of refraction to point $C .$ Since, the angle of refraction there is the same as the angle of incidence at $A, n \sin θ_{3} = \sin θ_{1} .$ Now, $α + β = 120 °, α = 90 ° - θ_{3}$ and $β = 90 ° - θ_{2}$ , as before. This means $θ_{2} + θ_{3} = 60 ° .$ Thus, the law of refraction leads to

$\sin θ_{1} = n \sin (60 ° - θ_{2}) \Rightarrow \sin θ_{1} = n \sin 60 ° \cos θ_{2} - n \cos 60 ° \sin θ_{2}$

where the trigonometric identity

$\sin (A - B) = \sin A \cos B - \cos A \sin B$

is used. Next, we apply the law of refraction to point $A :$

$\sin θ_{1} = n \sin θ_{2} \Rightarrow \sin θ_{2} = (1 / n) \sin θ_{1}$

which yields $\cos θ_{2} = \sqrt{1 - \sin^{2} θ_{2}} = \sqrt{1 - c 1 / n^{2} h \sin^{2} θ_{1}}$ . Thus,

$\sin θ_{1} = n \sin 60 ° \sqrt{1 - b 1 / n g^{2} \sin^{2} θ_{1}} - \cos 60 ° \sin θ_{1}$

or $b 1 + \cos 60 ° g \sin θ_{1} = \sin 60 ° \sqrt{n^{2} - \sin^{2} θ_{1}}$

Squaring both sides and solving for $\sin θ_{1},$ we obtain

$\sin θ_{1} = \frac{n \sin 60 °}{\sqrt{b 1 + \cos 60 ° g^{2} + \sin^{2} 60 °}} = \frac{1.60 \sin 60 °}{\sqrt{b 1 + \cos 60 ° g^{2} + \sin^{2} 60 °}} = 0.80$

and $θ_{1} = 53.1 °$ .

Important Questions on Electromagnetic Waves

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