Problem Analysis & Geometry

The problem involves a monochromatic light beam incident on a regular hexagonal glass prism. The beam is symmetric about the horizontal axis and falls on the top-left and bottom-left faces. We are given:

• Prism Shape: Regular Hexagon.
• Beam Width: Extends from the central axis up to the middle of the adjacent inclined faces.
• Condition: Two separate parallel beams emerge from the opposite side.

For a clear beam to emerge separate and parallel, the “limiting ray” (the ray entering at the extreme edge of the beam, i.e., at the midpoint of the entry face) must critically hit the edge of the exit face. If it hits beyond the edge, it would strike a different face (the bottom horizontal face) and likely undergo total internal reflection or emerge in a different direction, disrupting the parallel beam structure.

Step 1: Angle of Incidence

Consider the top-left face of the hexagon. In a standard orientation where the top and bottom faces are horizontal:

• The angle of the top-left face with the horizontal is $60^\circ$.
• The normal to this face makes an angle of $30^\circ$ with the horizontal.
• Since the incident ray is horizontal, the angle of incidence $i$ is the angle between the ray and the normal.

$$i = 30^\circ$$

Using Snell’s Law at the first interface ($n_1 = 1, n_2 = \mu$):

$$1 \cdot \sin(30^\circ) = \mu \sin(r)$$ $$\frac{1}{2} = \mu \sin(r) \quad \dots \text{(Equation 1)}$$

Step 2: Geometric Condition for Critical Ray

The beam enters the prism covering the section from the vertex (on the axis) to the midpoint $M$ of the face. For the emerging beam to be formed cleanly and exit through the opposite parallel face (the bottom-right face), the limiting ray (entering at $M$) must hit the prism boundary exactly at the vertex $V$ (the rightmost vertex on the symmetry axis).

We can determine the angle of deviation using coordinate geometry. Let the side length of the hexagon be $a$.

• Coordinate of Entry Point ($M$): This is the midpoint of the top-left side.
  Coordinates: $x_1 = -\frac{3a}{4}, \quad y_1 = \frac{\sqrt{3}a}{4}$ (relative to center at 0,0).
• Coordinate of Target Vertex ($V$): This is the rightmost vertex of the hexagon.
  Coordinates: $x_2 = a, \quad y_2 = 0$.

The slope of the refracted ray traveling from $M$ to $V$ is:

$$\tan(\alpha) = \left| \frac{y_2 – y_1}{x_2 – x_1} \right| = \frac{\frac{\sqrt{3}a}{4} – 0}{a – (-\frac{3a}{4})} = \frac{\frac{\sqrt{3}a}{4}}{\frac{7a}{4}} = \frac{\sqrt{3}}{7}$$

Here, $\alpha$ is the angle the refracted ray makes with the horizontal axis.

Step 3: Calculating Refractive Index ($\mu$)

From the geometry, the refracted ray bends downwards from the horizontal. The angle with the normal is $r$. The normal is at $30^\circ$ to the horizontal. Therefore, the angle the refracted ray makes with the horizontal is:

$$\alpha = 30^\circ – r$$

Substituting the value of $\tan(\alpha)$ we found:

$$\tan(30^\circ – r) = \frac{\sqrt{3}}{7}$$

Now we expand $\tan(30^\circ – r)$ using the formula $\tan(A-B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$:

$$\frac{\tan 30^\circ – \tan r}{1 + \tan 30^\circ \tan r} = \frac{\sqrt{3}}{7}$$

Substitute $\tan 30^\circ = \frac{1}{\sqrt{3}}$:

$$\frac{\frac{1}{\sqrt{3}} – \tan r}{1 + \frac{1}{\sqrt{3}} \tan r} = \frac{\sqrt{3}}{7}$$

Multiply numerator and denominator of LHS by $\sqrt{3}$:

$$\frac{1 – \sqrt{3}\tan r}{\sqrt{3} + \tan r} = \frac{\sqrt{3}}{7}$$

Cross-multiply:

$$7(1 – \sqrt{3}\tan r) = \sqrt{3}(\sqrt{3} + \tan r)$$ $$7 – 7\sqrt{3}\tan r = 3 + \sqrt{3}\tan r$$ $$4 = 8\sqrt{3}\tan r$$ $$\tan r = \frac{4}{8\sqrt{3}} = \frac{1}{2\sqrt{3}}$$

Now we need $\sin r$ to use in Snell’s Law. We use the identity $\sin r = \frac{\tan r}{\sqrt{1 + \tan^2 r}}$:

$$\sin r = \frac{\frac{1}{2\sqrt{3}}}{\sqrt{1 + \left(\frac{1}{2\sqrt{3}}\right)^2}} = \frac{\frac{1}{2\sqrt{3}}}{\sqrt{1 + \frac{1}{12}}} = \frac{\frac{1}{2\sqrt{3}}}{\sqrt{\frac{13}{12}}} = \frac{1}{\sqrt{13}}$$

Finally, substitute $\sin r$ back into Equation 1:

$$\mu = \frac{\sin 30^\circ}{\sin r} = \frac{1/2}{1/\sqrt{13}} = \frac{\sqrt{13}}{2}$$

Conclusion

The required refractive index of the glass for two separate parallel beams to emerge is:

$$ \mu = \frac{\sqrt{13}}{2} $$

The answer does not match that given in the book.