suppose that the angle of incidence is 1 = 37.1°, the thickness of the pane is 4.17 mm, and the refractive index of the pane is n2 = 1.50. Find the amount (in mm) by which the emergent ray is displaced relative to the incident ray.
Refer to the figure at
http://img361.imageshack.us/img361/773/displacementduetorefracra3.gif
Use the formula derived at
http://www.kwantlen.bc.ca/science/physics/faculty/mcoombes/P1102_Solutions/Refraction/P1102_01_Solutions.htm
but use the appropriate values of the incidence angle, refractive index and the thickness.
To calculate the displacement of the emergent ray relative to the incident ray, we need to use Snell's Law and basic trigonometry.
1. Snell's Law states that n1 * sin(theta1) = n2 * sin(theta2), where n1 and n2 are the refractive indices of the incident and refracted mediums, respectively, and theta1 and theta2 are the angles of incidence and refraction, respectively.
2. We need to start by finding the angle of refraction (theta2). Rearranging Snell's Law, we have sin(theta2) = (n1 / n2) * sin(theta1). Plugging in the given values, we get sin(theta2) = (1.00 / 1.50) * sin(37.1°). Evaluating this using a calculator, sin(theta2) ≈ 0.494.
3. Next, we can find the angle of refraction (theta2) using the inverse sine function: theta2 = sin^(-1)(sin(theta2)). Plugging in the value, we get theta2 ≈ 29.2°.
4. Now, we can use basic trigonometry to find the displacement of the emergent ray. The displacement is the difference in the horizontal position between the incident and emergent rays. We can use the equation: displacement = (thickness of the pane) * tan(theta2). Plugging in the given thickness of the pane (4.17 mm) and the value of theta2 in radians (approximately 0.5084), we get the displacement ≈ 4.17 mm * tan(0.5084). Evaluating this using a calculator, we get displacement ≈ 1.84 mm.
Therefore, the emergent ray is displaced approximately 1.84 mm relative to the incident ray.