Questions: Internal Reflection Light A ray in glass (n=1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract. (Water n=1.33, Air n=1.00)

$Internal Reflection Light A ray in glass (n=1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract. (Water n=1.33, Air n=1.00)$

Transcript text: Internal Reflection Light A ray in glass ( $n=1.51$ ) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract. (Water $n=1.33$, Air $n=1.00$ )

Solution

Solution Steps

Step 1: Determine the Critical Angle

To determine whether the light ray reflects internally or refracts into the air, we need to calculate the critical angle for the glass-air boundary. The critical angle ($\theta_c$) can be found using Snell's Law:

\[ n_{\text{glass}} \cdot \sin(\theta_c) = n_{\text{air}} \cdot \sin(90^\circ) \]

Since $\sin(90^\circ) = 1$, the equation simplifies to:

\[ \sin(\theta_c) = \frac{n_{\text{air}}}{n_{\text{glass}}} \]

Substituting the given values:

\[ \sin(\theta_c) = \frac{1.00}{1.51} \]

\[ \sin(\theta_c) \approx 0.6623 \]

Now, calculate $\theta_c$:

\[ \theta_c = \arcsin(0.6623) \approx 41.1^\circ \]

Step 2: Compare the Incident Angle with the Critical Angle

The incident angle given is $49.2^\circ$. We compare this with the critical angle:

If the incident angle is greater than the critical angle, the light will reflect internally.
If the incident angle is less than or equal to the critical angle, the light will refract into the air.

Since $49.2^\circ > 41.1^\circ$, the light will reflect internally.