Solution: Light Propagation in a Tapered Glass Plate
Method: The “Unfolding” Technique
Instead of tracing the zigzag path of the light ray through the tapered plate, we can “unfold” the reflections. By mirroring the wedge repeatedly, the optical path becomes a straight line passing through a set of fan-like sectors. This simplifies the geometry significantly.
Geometric Analysis:
Let the length of the plate be $l$, the entrance thickness be $d$, and the taper angle be $\beta$ (where $\beta \ll 1^\circ$).
The plate forms a section of a circular sector with the vertex at $O$. The radius of the entrance face is:
$$ R \approx \frac{d}{\beta} $$
The radius of the exit face is:
$$ R’ = R – l = \frac{d}{\beta} – l $$
Consider a light ray entering the center of the input face at an angle $\alpha$ with the central axis. In the “unfolded” straight-line geometry, the ray travels in a straight line starting from distance $R$. The condition for the light to pass through the plate is that this straight line must intersect the “exit” arc at radius $R’$.
If the ray angle $\alpha$ is too large, the ray will miss the inner circle (radius $R’$) entirely. Physically, this corresponds to the ray turning back due to multiple reflections before reaching the end.
Limiting Condition:
The ray is tangent to the inner circle of radius $R’$. From the geometry of the triangle formed by the vertex $O$, the entry point, and the tangent point:
$$ d_{\text{min}} = R \sin \alpha $$
where $d_{\text{min}}$ is the distance of closest approach to the vertex. For the ray to pass through (intersect the exit face), we must have:
$$ d_{\text{min}} \le R’ $$
$$ R \sin \alpha \le R – l $$
$$ \sin \alpha \le \frac{R – l}{R} $$
$$ \sin \alpha \le 1 – \frac{l}{R} $$
Substituting $R = d/\beta$:
$$ \sin \alpha \le 1 – \frac{l}{d/\beta} $$
$$ \sin \alpha \le 1 – \frac{\beta l}{d} $$
Thus, the range of incidence angles $\alpha$ is:
$$ 0^\circ \le \alpha \le \sin^{-1} \left( 1 – \frac{\beta l}{d} \right) $$
Method: The “Unfolding” Technique
Instead of tracing the zigzag path of the light ray through the tapered plate, we can “unfold” the reflections. By mirroring the wedge repeatedly, the optical path becomes a straight line passing through a set of fan-like sectors. This simplifies the geometry significantly.
Geometric Analysis:
Let the length of the plate be $l$, the entrance thickness be $d$, and the taper angle be $\beta$ (where $\beta \ll 1^\circ$).
The plate forms a section of a circular sector with the vertex at $O$. The radius of the entrance face is:
$$ R \approx \frac{d}{\beta} $$The radius of the exit face is:
$$ R’ = R – l = \frac{d}{\beta} – l $$Consider a light ray entering the center of the input face at an angle $\alpha$ with the central axis. In the “unfolded” straight-line geometry, the ray travels in a straight line starting from distance $R$. The condition for the light to pass through the plate is that this straight line must intersect the “exit” arc at radius $R’$.
If the ray angle $\alpha$ is too large, the ray will miss the inner circle (radius $R’$) entirely. Physically, this corresponds to the ray turning back due to multiple reflections before reaching the end.
Limiting Condition:
The ray is tangent to the inner circle of radius $R’$. From the geometry of the triangle formed by the vertex $O$, the entry point, and the tangent point:
$$ d_{\text{min}} = R \sin \alpha $$where $d_{\text{min}}$ is the distance of closest approach to the vertex. For the ray to pass through (intersect the exit face), we must have:
$$ d_{\text{min}} \le R’ $$ $$ R \sin \alpha \le R – l $$ $$ \sin \alpha \le \frac{R – l}{R} $$ $$ \sin \alpha \le 1 – \frac{l}{R} $$Substituting $R = d/\beta$:
$$ \sin \alpha \le 1 – \frac{l}{d/\beta} $$ $$ \sin \alpha \le 1 – \frac{\beta l}{d} $$Thus, the range of incidence angles $\alpha$ is: