Question

When light passes from a more-dense to a less-dense medium—from glass to air, for example—the angle of refraction predicted by Snell's Law can be 90° or larger. In this case the light beam is actually reflected back into the denser medium. This phenomenon, called total internal reflection, is the principle behind fiber optics, Snell's Law is given below. sin(81). Vi sin(02) Vz Set B = 90° in Snell's Law and solve for A, to determine the critical angle of incidence at which total internal reflection begins to occur when light passes from diamond to air. (Note that the index of refraction from diamond to alr is the reciprocal of the index from air to diamond. Round your answer to one decimal place.) P1 = 155.4 x • Refraction O1 = 0.4 x • from air Substance to substance Water 1.33 Alcohol 1.36 Glass 1.52 Diamond 2.41

Answer 1

76.78°

Explanation:

See picture attached

According to Snell's law

`\bf \displaystyle(sin(\theta_2))/(sin(\theta_2))=\displaystyle(n_1)/(n_2)`

The critical angle of incidence
`\bf \theta_1` occurs when
`\bf \theta_2=90^(o)`

The index of refraction of air can be considered 1. The index of refraction of diamond is 2.417

Solving for
`\bf \theta_1`

`\bf \displaystyle(sin(90^(o)))/(sin(\theta_1))=\displaystyle(2.417)/(1)\Rightarrow sin(\theta_1)=\displaystyle(1)/(2.417)\Rightarrow sin(\theta_1)=0.41374\Rightarrow\\\\\Rightarrow \theta_1=arcsin(0.41374)\Rightarrow \theta_1=0.42656\;rad\Rightarrow \boxed{\theta_1=76.78^(o)}`