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You send a beam of light from a material with index of refraction 1.25 1.25 into an unknown material. In order to help identify this material, you determine its index of refraction by measuring the angles of incidence and refraction for which you find the values 40.1 ∘ 40.1∘ and 36.9 ∘ , 36.9∘, respectively. What is the index of refraction n n of the unknown material?

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sqdancefan

Answer:

  1.34

Step-by-step explanation:

Snell's law tells you ...

  [tex]n_i\cdot\sin{\theta_i}=n_r\cdot\sin{\theta_r}[/tex]

Solving for [tex]n_r[/tex], we get ...

  [tex]n_r=n_i\cdot\dfrac{\sin{\theta_i}}{\sin{\theta_r}}=1.25\cdot\dfrac{\sin{40.1^\circ}}{\sin{36.9^\circ}}\approx 1.34[/tex]

The unknown material has an index of refraction of 1.34.

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