Understanding Reflection from a Plane Mirror

When a ray of light is incident on a plane mirror, it follows the laws of reflection. These laws state that:

• The angle of incidence is equal to the angle of reflection.
• The incident ray, the reflected ray, and the normal to the mirror at the point of incidence all lie in the same plane.

The angle of incidence (\(i\)) is defined as the angle between the incident ray and the normal (a line perpendicular) to the mirror surface at the point where the ray hits the mirror.

Using Direction Vectors to Represent Rays

The direction of the incident ray is given by the vector \(\vec{I} = \frac{1}{2} (\hat i + \sqrt 3 \hat j) = \frac{1}{2} \hat i + \frac{\sqrt 3}{2} \hat j\).

The direction of the reflected ray is given by the vector \(\vec{R} = \frac{1}{2} (\hat i - \sqrt 3 \hat j) = \frac{1}{2} \hat i - \frac{\sqrt 3}{2} \hat j\).

The magnitude of the incident ray vector is \(|\vec{I}| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\).

The magnitude of the reflected ray vector is \(|\vec{R}| = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt 3}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1\).

Determining the Normal Vector Direction

The normal vector (\(\hat{n}\)) to the mirror surface is perpendicular to the mirror. Geometrically, the normal vector bisects the angle between the incident ray direction and the reverse of the reflected ray direction, or equivalently, the normal direction is perpendicular to the vector that bisects the angle between the incident and reflected rays. A simpler approach for reflection from a plane mirror is that the normal vector is parallel to the difference between the reflected and incident ray vectors, i.e., proportional to \(\vec{R} - \vec{I}\).

Let's compute the difference vector:

\(\vec{R} - \vec{I} = (\frac{1}{2} \hat i - \frac{\sqrt 3}{2} \hat j) - (\frac{1}{2} \hat i + \frac{\sqrt 3}{2} \hat j)\)

\(\vec{R} - \vec{I} = (\frac{1}{2} - \frac{1}{2}) \hat i + (-\frac{\sqrt 3}{2} - \frac{\sqrt 3}{2}) \hat j\)

\(\vec{R} - \vec{I} = 0 \hat i - \sqrt 3 \hat j = -\sqrt 3 \hat j\)

This indicates that the normal to the mirror surface is along the direction of \(\hat{j}\) or \(-\hat{j}\).

Looking at the incident ray vector \(\vec{I} = \frac{1}{2} \hat i + \frac{\sqrt 3}{2} \hat j\), its \(\hat{j}\) component is positive. The reflected ray vector \(\vec{R} = \frac{1}{2} \hat i - \frac{\sqrt 3}{2} \hat j\) has a negative \(\hat{j}\) component. If we consider the y-axis parallel to \(\hat{j}\), the incident ray is coming from the positive y direction and the reflected ray is going into the negative y direction. This implies the mirror is likely horizontal (parallel to the x-axis, \(\hat{i}\)) and the normal pointing outwards from the mirror surface into the region the ray is coming from is along \(+\hat{j}\). So, we take the unit normal vector as \(\hat{n} = \hat{j}\).

Calculating the Angle of Incidence

The angle of incidence \(i\) is the angle between the incident ray vector \(\vec{I}\) and the normal vector \(\hat{n}\). We can find this angle using the dot product formula:

\(\cos i = \frac{\vec{I} \cdot \hat{n}}{|\vec{I}| |\hat{n}|}\)

Calculate the dot product \(\vec{I} \cdot \hat{n}\):

\(\vec{I} \cdot \hat{n} = (\frac{1}{2} \hat i + \frac{\sqrt 3}{2} \hat j) \cdot \hat j\)

\(\vec{I} \cdot \hat{n} = (\frac{1}{2})(0) + (\frac{\sqrt 3}{2})(1) = \frac{\sqrt 3}{2}\)

We already know \(|\vec{I}| = 1\). The magnitude of the unit normal vector is \(|\hat{n}| = |\hat{j}| = 1\).

Substitute these values into the cosine formula:

\(\cos i = \frac{\frac{\sqrt 3}{2}}{(1)(1)} = \frac{\sqrt 3}{2}\)

Since \(i\) is the angle of incidence, it is an acute angle (between 0° and 90°). The angle whose cosine is \(\frac{\sqrt 3}{2}\) is 30°.

Therefore, the angle of incidence is \(i = 30^\circ\).

Conclusion

Using the direction vectors for the incident and reflected rays and determining the direction of the normal to the plane mirror, we calculated the angle between the incident ray and the normal using the dot product. This angle, which is the angle of incidence, is found to be 30°.