The refracting angle of a glass prism is 30°. A ray is incident onto one of the faces perpendicular to it. Find the angle δ between the incident ray and the ray that leaves the pri... The refracting angle of a glass prism is 30°. A ray is incident onto one of the faces perpendicular to it. Find the angle δ between the incident ray and the ray that leaves the prism.
Understand the Problem
The question is asking to find the angle δ between the incident ray and the ray that exits a glass prism, given certain parameters like the refracting angle and the refractive index of the glass.
Answer
The angle between the incident ray and the ray that leaves the prism is $48.6^\circ$.
Answer for screen readers
The angle $\delta$ between the incident ray and the ray that leaves the prism is approximately:
$$ \delta \approx 48.6^\circ $$
Steps to Solve
- Identify the parameters given in the problem
The refracting angle of the prism, $A = 30^\circ$, and the refractive index of the glass, $n = 1.5$, are given. The incident ray hits the face of the prism perpendicularly, which means the angle of incidence, $i = 0^\circ$.
- Use Snell's Law at the first interface
Since the ray is incident perpendicularly, we can apply Snell's Law at the air-glass boundary: $$ n_{\text{air}} \sin i = n_{\text{glass}} \sin r_1 $$ Here, $n_{\text{air}} = 1$, $i = 0^\circ$, and $r_1$ is the angle of refraction in the glass. We find: $$ 1 \cdot \sin(0) = 1.5 \cdot \sin(r_1) $$
Since $\sin(0) = 0$, it implies: $$ r_1 = 0^\circ $$
- Determine the angle of refraction at the second surface
Inside the prism, the ray will travel at an angle to the second face. The angle of incidence at the second surface is: $$ i_2 = A - r_1 = 30^\circ - 0^\circ = 30^\circ $$
Now, apply Snell's Law for the second surface: $$ n_{\text{glass}} \sin i_2 = n_{\text{air}} \sin r_2 $$ Substituting the known values: $$ 1.5 \cdot \sin(30^\circ) = 1 \cdot \sin r_2 $$
Calculating $\sin(30^\circ)$: $$ \sin(30^\circ) = 0.5 $$ So, we find: $$ 1.5 \cdot 0.5 = \sin r_2 $$ $$ \sin r_2 = 0.75 $$
- Calculate the angle $r_2$
To find $r_2$, we use the inverse sine: $$ r_2 = \sin^{-1}(0.75) $$
Calculating this gives: $$ r_2 \approx 48.6^\circ $$
- Find the angle $\delta$
Finally, the angle $\delta$ between the incident ray and the ray that exits the prism is given by: $$ \delta = r_2 \text{ (since the incident ray is at } 0^\circ) $$ Thus, $$ \delta = 48.6^\circ $$
The angle $\delta$ between the incident ray and the ray that leaves the prism is approximately:
$$ \delta \approx 48.6^\circ $$
More Information
The calculated angle corresponds closely to one of the options provided in the problem. The angle is determined using the principles of refraction and Snell's Law. A refracting angle of 30° combined with the refractive index gives a clear path for the outgoing ray.
Tips
- Confusing angles: It’s vital to clarify which angle refers to the incident ray and which refers to the refracting angles.
- Misapplication of Snell's Law: Remember to use the correct indices and make sure to arrange them appropriately based on the interface.
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