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)

Solution Steps

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 $\theta_c$:

$$\theta_c = \arcsin(0.6623) \approx 41.1^\circ$$

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.

Final Answer