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Light rays in a material with index of refraction 1.31 can undergo total internal reflection when they strike the interface with another material at a critical angle of incidence. Find this material\'s index of refraction when the required critical angle is 78.3°.

Answer :

Answer:

The refractive index of the material is 1.28.

Explanation:

It is given that,

Refractive index of medium 1, n₁ = 1.31

Critical angle, [tex]\theta_c=78.3[/tex]

At critical angle rays will reflects at 90 degrees. Using Snell's law as:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

At critical angle, [tex]n_1sin\ \theta_c=n_2\ sin90[/tex]

[tex]n_1sin\ \theta_c=n_2[/tex]

[tex]n_2=1.31\times sin(78.3)[/tex]

[tex]n_2=1.28[/tex]

So, the refractive index of the material is 1.28. Hence, this is the required solution.

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