In Snell's law, sin(theta1)/sin(theta2) equals which ratio?

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Multiple Choice

In Snell's law, sin(theta1)/sin(theta2) equals which ratio?

Explanation:
The main idea is that refraction is governed by Snell’s law, which links the angles to the refractive indices of the two media. Snell’s law states that n1 sinθ1 = n2 sinθ2, so the ratio of the sines is sinθ1 / sinθ2 = n2 / n1. Now connect that to wavelength. The wavelength in a medium is shorter by a factor of the refractive index: lambda_i = lambda0 / n_i. Therefore, the ratio of wavelengths is lambda1 / lambda2 = (lambda0/n1) / (lambda0/n2) = n2 / n1. Putting these together shows sinθ1 / sinθ2 = lambda1 / lambda2. That’s why the ratio of the sines matches the ratio of the wavelengths, making lambda1/lambda2 the correct expression.

The main idea is that refraction is governed by Snell’s law, which links the angles to the refractive indices of the two media. Snell’s law states that n1 sinθ1 = n2 sinθ2, so the ratio of the sines is sinθ1 / sinθ2 = n2 / n1.

Now connect that to wavelength. The wavelength in a medium is shorter by a factor of the refractive index: lambda_i = lambda0 / n_i. Therefore, the ratio of wavelengths is lambda1 / lambda2 = (lambda0/n1) / (lambda0/n2) = n2 / n1.

Putting these together shows sinθ1 / sinθ2 = lambda1 / lambda2. That’s why the ratio of the sines matches the ratio of the wavelengths, making lambda1/lambda2 the correct expression.

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