Two light waves that have wavelengths of 700 and \(400 \mathrm{~nm}\) enter a block of glass (from air) with the same angle of incidence. Which has the larger angle of refraction? Why? Would the answer be different if the light waves were going from glass into air?

Snell's Law relates the angles of incidence and refraction with the refractive indices of the two media involved. The formula for Snell's Law is given by: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\) where \(n_1\) and \(n_2\) are the refractive indices of the first and second medium, respectively, and \(\theta_1\) and \(\theta_2\) are the angle of incidence and angle of refraction, respectively. The refractive index of a medium also depends on the wavelength of the light passing through it. This property is known as dispersion. A higher wavelength experiences lower refraction index.

Assume the angle of incidence for both light waves is \(\theta_1\). Let \(n_1\) be the refractive index of air, \(n_2^{700}\) and \(n_2^{400}\) be the refractive indices for 700 nm and 400 nm light waves in glass, respectively. Since both light waves have the same angle of incidence and a higher wavelength means a lower refraction index in the glass, the 700 nm light wave will have a smaller \(n_2^{700}\) than \(n_2^{400}\). Using Snell's Law, \(n_1 \sin(\theta_1) = n_2^{700} \sin(\theta_{2}^{700})\) and \(n_1 \sin(\theta_1) = n_2^{400} \sin(\theta_{2}^{400})\) Since \(n_2^{700} < n_2^{400}\) and \(\sin(\theta)\) is an increasing function, \(\theta_{2}^{700} > \theta_{2}^{400}\). Thus, the 700 nm light wave will have a larger angle of refraction when entering the glass from air.

Now, let's analyze the case when the light waves pass from the glass back into the air. The refractive indices will be reversed (\(n_1^{700}\), \(n_1^{400}\) in glass and \(n_2\) in air). Since the 700 nm light wave has a smaller refractive index, it will have a smaller \(n_1^{700}\) than \(n_1^{400}\). Using Snell's Law again, \(n_1^{700} \sin(\theta_1^{700}) = n_2 \sin(\theta_{2}^{700})\) and \(n_1^{400} \sin(\theta_1^{400}) = n_2 \sin(\theta_{2}^{400})\) As \(n_1^{700} < n_1^{400}\) and \(\sin(\theta)\) is an increasing function, \(\theta_{2}^{700} > \theta_{2}^{400}\). In conclusion, the 700 nm light wave will have a larger angle of refraction both when entering the glass from air and when passing from the glass back into the air.