To determine the refractive index of the medium where light undergoes total internal reflection at an angle of incidence of 45°, we need to understand the principle of total internal reflection.
Total internal reflection occurs when light travels from a denser medium to a rarer medium and the angle of incidence is greater than the critical angle. The refractive index relationship is given by:
\(n_{\text{medium}} \sin \theta_c = n_{\text{air}} \sin 90^\circ\)
Where \( n_{\text{medium}} \) is the refractive index of the medium, \( \theta_c \) is the critical angle, and \( n_{\text{air}} \) is the refractive index of air, which is approximately 1. Since sin 90° is 1, the equation simplifies to:
\(n_{\text{medium}} \sin \theta_c = 1\)
Given that the angle of incidence in the medium is 45° and it is equal to the critical angle for total internal reflection, we have:
\(n_{\text{medium}} \sin 45^\circ = 1\)
We know that \(\sin 45^\circ = \frac{\sqrt{2}}{2}\), so substituting it gives us:
\(n_{\text{medium}} \times \frac{\sqrt{2}}{2} = 1\)
Solving for \( n_{\text{medium}} \), we find:
\(n_{\text{medium}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414\)
The option closest to this value is:
The value of 1.5 is the closest to the calculated refractive index of approximately 1.414. Thus, the correct answer is 1.5.