91Ó°ÊÓ

We will use Snell's Law, which is given by: \[n_1\sin\theta_1 = n_2\sin\theta_2\] Where, \(n_1\) is the refractive index of the medium through which the light is incident (air), \(n_2\) is the refractive index of the medium through which the light is refracted (glass), \(\theta_1\) is the angle of incidence, and \(\theta_2\) is the angle of refraction. Since the refractive index of air is \(1\), we can find the angle of refraction, \(\theta_2\) as follows: \[\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2}\]

Now, let's calculate the angle of refraction \(\theta_2\) using the values given in the problem: wavelength of light in air = 60º refractive index of glass, \(\mu\) = \(3/2\) \[\sin\theta_2 = \frac{\sin 60^\circ}{\frac{3}{2}} = \frac{1}{2(\frac{3}{2})} = \frac{1}{3}\] To obtain the angle of refraction, we can take the arcsin of \(\sin\theta_2\): \[\theta_2 = \arcsin\frac{1}{3}\]

Consider the incident and refracted beams as right-angled triangles with height \(h\) and base length equal to half the diameter (\(d/2\)). Now, apply the tangent function to the incident and refracted beams: For the incident beam, \[\tan 60^\circ = \frac{h}{d_1/2}\] As for the refracted beam, \[\tan \theta_2 = \frac{h}{d_2/2}\] Now, divide the equation of the refracted beam by the incident beam equation: \[\frac{\tan \theta_2}{\tan 60^\circ} = \frac{d_1/2}{d_2/2}\] Plug in the values, \(\tan 60^\circ\) and \(\tan \theta_2\), as well as the diameter of the incident beam, \(d_1 = 2 \text{ cm}\), into the equation: \[\frac{\tan (\arcsin(1/3))}{\tan 60^\circ} = \frac{2}{d_2}\] Solve for the diameter of the refracted beam (\(d_2\)): \[d_2 = 2 \cdot \frac{\tan 60^\circ}{\tan (\arcsin(1/3))}\]

All we need to do now is evaluate the expression and round the result to a reasonable accuracy: \[d_2 = 2 \cdot \frac{\tan 60^\circ}{\tan (\arcsin(1/3))} \approx 3.26 \text{ cm}\] Hence, the diameter of the refracted beam is \(3.26 \text{ cm}\) (option C).