3.flashlight beam strikes the surface of a pane of glass (n=1.56) at a 67 angle to the normal .What is the angle of refraction [1]?

To find the angle of refraction, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. Snell's law is given by:<br /><br />\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]<br /><br />where:<br />- \( n_1 \) is the refractive index of the first medium (air in this case, which is approximately 1),<br />- \( \theta_1 \) is the angle of incidence,<br />- \( n_2 \) is the refractive index of the second medium (glass in this case, which is 1.56),<br />- \( \theta_2 \) is the angle of refraction.<br /><br />Given:<br />- \( n_1 = 1 \) (for air),<br />- \( \theta_1 = 67^\circ \) (angle of incidence),<br />- \( n_2 = 1.56 \) (for glass).<br /><br />We need to find \( \theta_2 \), the angle of refraction.<br /><br />Using Snell's law:<br /><br />\[ 1 \cdot \sin(67^\circ) = 1.56 \cdot \sin(\theta_2) \]<br /><br />First, calculate \( \sin(67^\circ) \):<br /><br />\[ \sin(67^\circ) \approx 0.9205 \]<br /><br />Now, substitute this value into the equation:<br /><br />\[ 0.9205 = 1.56 \cdot \sin(\theta_2) \]<br /><br />Solve for \( \sin(\theta_2) \):<br /><br />\[ \sin(\theta_2) = \frac{0.9205}{1.56} \]<br /><br />\[ \sin(\theta_2) \approx 0.591 \]<br /><br />Finally, find \( \theta_2 \) by taking the inverse sine (arcsin) of 0.591:<br /><br />\[ \theta_2 \approx \arcsin(0.591) \]<br /><br />Using a calculator:<br /><br />\[ \theta_2 \approx 36.1^\circ \]<br /><br />Therefore, the angle of refraction is approximately \( 36.1^\circ \).