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What is the index of refraction of a refractive medium if the angle of incidence in the air is 40 and the angle of refraction is 29?.

Sagot :

(1) sin 40° = n(sin 200⁰).

1. A ray of light traveling from air into crown glass strikes the surface at an angle of 30°.

theleangreenbean

We can often use Snell's Law to help us find the index of refraction of a refractive medium.

Snell's Law:  [tex]\dfrac{n_1}{n_2}=\dfrac{\sin(\theta_2)}{\sin(\theta_1)}[/tex]

  • [tex]n_1[/tex] = the refractive index of the first medium
  • [tex]n_2[/tex] = the refractive index of the second medium
  • [tex]\theta_1[/tex] = the angle of incidence
  • [tex]\theta_2[/tex] = the angle of refraction

In this question, we're given the following:

  • The angle of incidence is [tex]40^\circ[/tex] ⇒ [tex]\theta_1=40^\circ[/tex]
  • The first medium is air, which has a refractive index of 1.0003 ⇒[tex]n_1 = 1.0003[/tex]
  • The angle of refraction is [tex]29^\circ[/tex] ⇒ [tex]\theta_2=29^\circ[/tex]
  • Solve for [tex]n_2[/tex].

Since we know we're solving for the index of refraction of the second medium, isolate [tex]n_2[/tex] in Snell's Law:

[tex]\dfrac{n_1}{n_2}=\dfrac{\sin(\theta_2)}{\sin(\theta_1)}\\\\\\\dfrac{n_2}{n_1}=\dfrac{\sin(\theta_1)}{\sin(\theta_2)}\\\\\\n_2=\dfrac{\sin(\theta_1)}{\sin(\theta_2)}*n_1[/tex]

Plug in all the information we know and find [tex]n_2[/tex]:

[tex]n_2=\dfrac{\sin(40^\circ)}{\sin(29^\circ)}*1.0003\\\\\\n_2\approx1.3263[/tex]

Answer

Therefore, the index of refraction of the refractive medium is approximately 1.3263.

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