6. A light ray travels through the water (n1 =1.33) at theta = 15° until it enters a diamond ring(n2=2.42)a. What is the angle of the refracted light?|b. Given the index of refraction for diamond and the speed of light constant, what isthe velocity of light in diamond?

Respuesta :

a.

In order to calculate the angle of refraction, we can use the law of refraction below:

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

Where n1 and n2 are the index of refraction and theta1 and theta2 are the incident and refracted angles.

So, using n1 = 1.33, theta1 = 15° and n2 = 2.42, we have:

[tex]\begin{gathered} 1.33*sin\left(15°\right)=2.42*sin\theta_2 \\ 1.33*0.258819=2.42*sin\theta_2 \\ 0.34422927=2.42*sin\theta_2 \\ sin\theta_2=\frac{0.34422927}{2.42} \\ sin\theta_2=0.1422435 \\ \theta_2=8.18° \end{gathered}[/tex]

b.

To calculate the velocity of light in diamond, we can use the formula below:

[tex]\begin{gathered} n=\frac{c}{v} \\ 2.42=\frac{3*10^8}{v} \\ v=\frac{3*10^8}{2.42} \\ v=1.24*10^8\text{ m/s} \end{gathered}[/tex]

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