A glass slab of refractive index \(1\cdot5\) is cut in the shape of a parallelepiped \(ABCD\) as shown in the figure. Surface \(BC\) is polished to form a perfect reflector. Light is incident at surface \(AB\) from inside at an angle \(\pi/3\). Which one of the following is the correct angle \(ABC\) (\(\theta\)) such that the reflected light retraces its path?
To solve this problem, we need to determine the angle \( \theta \) such that the reflected light retraces its path after hitting the surface \( BC \) which is polished to form a perfect reflector. The light is initially incident on surface \( AB \) at an angle of \( \pi/3 \) from inside the glass slab.
Given:
- Refractive index of glass, \( n = 1.5 \)
- Incident angle at \( AB \), \( i = \pi/3 \)
The condition for the light to retrace its path is that after reflection at \( BC \), the light must strike \( AB \) normally, so all the angles in the path must add up such that the system behaves as a retro-reflector.
Let's denote:
- \( \angle ABC = \theta \)
1. Snell's Law at \( AB \):
\(n \cdot \sin(i) = \sin(r)\), where \( r \) is the angle of refraction at \( AB \).
Substituting the known values:
\(1.5 \cdot \sin(\pi/3) = \sin(r)\)
\(\sin(\pi/3) = \frac{\sqrt{3}}{2}\), hence:
\(1.5 \cdot \frac{\sqrt{3}}{2} = \sin(r)\)
\(\sin(r) = \frac{3\sqrt{3}}{4}\)
Since \( \sin(r) \) must be \( \leq 1 \), \( \pi/3 \) complies with potential angles for refraction.
2. Reflection Condition at \( BC \):
For the light to retrace its path:
\(\pi - (r + \theta) = i\)
Since the light must retrace:
\(r + \theta = \pi - i\)
\(\theta = \pi - i - r\)
3. Substitute Known Values:
\(\theta = \pi - \frac{\pi}{3} - r\)
Given the angles, if we assume \( r = \frac{\pi}{3} \):
Therefore, the correct angle \( \angle ABC (\theta) \) is:
\(\pi/3\)
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