This problem involves optics, specifically the reflection and refraction of light at an interface between two media, and the concept of total internal reflection. We need to find the critical angle for the glass-air interface given specific conditions about a light ray incident from air onto glass.

Understanding the Perpendicular Rays Condition

The question states that a ray of light travels from air to a glass surface. The angle of incidence ($i$) is given as $60^\circ$. A crucial piece of information is that the reflected ray and the refracted ray are perpendicular to each other.

• Let the angle of incidence be $i$. Given $i = 60^\circ$.
• According to the law of reflection, the angle of reflection ($r_{refl}$) is equal to the angle of incidence. So, $r_{refl} = i = 60^\circ$.
• Let the angle of refraction be $r$.
• The reflected ray and the refracted ray are measured with respect to the normal to the surface. The angle between the reflected ray and the normal is $r_{refl}$, and the angle between the refracted ray and the normal is $r$.
• The condition that the reflected and refracted rays are perpendicular means the angle between them is $90^\circ$. This angle is the sum of the angle of reflection and the angle of refraction, measured from the normal: $r_{refl} + r = 90^\circ$.
• Substituting the value of $r_{refl}$: $60^\circ + r = 90^\circ$.
• Solving for $r$: $r = 90^\circ - 60^\circ = 30^\circ$.

This specific condition ($i + r = 90^\circ$) occurs when the angle of incidence is Brewster's angle, and it allows us to find the refractive index of the second medium.

Calculating Glass Refractive Index using Snell's Law

Now we use Snell's Law to relate the angles of incidence and refraction to the refractive indices of the two media (air and glass).

• Snell's Law is given by: $n_1 \sin(i) = n_2 \sin(r)$.
• Here, medium 1 is air and medium 2 is glass.
• The refractive index of air, $n_1$, is approximately $1$.
• The refractive index of glass, $n_2$, is what we need to find. Let's denote it as $n_{glass}$.
• The angle of incidence is $i = 60^\circ$.
• The angle of refraction is $r = 30^\circ$.
• Plugging these values into Snell's Law: $1 \times \sin(60^\circ) = n_{glass} \times \sin(30^\circ)$.
• We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ and $\sin(30^\circ) = \frac{1}{2}$.
• So, the equation becomes: $1 \times \frac{\sqrt{3}}{2} = n_{glass} \times \frac{1}{2}$.
• To find $n_{glass}$, we rearrange the equation: $n_{glass} = \frac{\sin(60^\circ)}{\sin(30^\circ)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$.
• $n_{glass} = \sqrt{3}$.

Therefore, the refractive index of the glass is $\sqrt{3}$.

Finding the Critical Angle for Glass-Air Interface

The question asks for the critical angle ($c$) for total internal reflection at the glass-air interface. Total internal reflection occurs when light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index).

• In this case, light travels from glass ($n_{glass} = \sqrt{3}$) to air ($n_{air} = 1$).
• The critical angle ($c$) is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^\circ$.
• We apply Snell's Law again, but this time for light going from glass to air: $n_{glass} \sin(c) = n_{air} \sin(90^\circ)$.
• We know $n_{glass} = \sqrt{3}$, $n_{air} = 1$, and $\sin(90^\circ) = 1$.
• Substituting these values: $\sqrt{3} \times \sin(c) = 1 \times 1$.
• $\sqrt{3} \sin(c) = 1$.
• Solving for $\sin(c)$: $\sin(c) = \frac{1}{\sqrt{3}}$.
• Taking the inverse sine to find the critical angle $c$: $c = \arcsin\left(\frac{1}{\sqrt{3}}\right)$.

This value matches one of the options provided.