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Light shines from a laser in air down into water. If the laser beam in air makes an angle of with the water's surface, what angle will it make with the surface under water?
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Identify the known values: the angle of incidence (angle in air) is 40 degrees, and the refractive indices for air and water are approximately 1.0 and 1.33, respectively.
Use Snell's Law to relate the angles and refractive indices: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of air and water, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
Substitute the known values into Snell's Law: \( 1.0 \sin(40^\circ) = 1.33 \sin(\theta_2) \).
Solve for \( \sin(\theta_2) \) by rearranging the equation: \( \sin(\theta_2) = \frac{1.0 \sin(40^\circ)}{1.33} \).
Calculate \( \theta_2 \) by taking the inverse sine (arcsin) of the result from the previous step to find the angle of refraction in water.