When a light ray strikes a surface separating two media with different optical properties, part of the light energy is reflected back into the medium from which it came. When light strikes a perfectly reflecting surface (e.g., a mirror), all the light energy is reflected.
Key terms:
The surface separating two media is called the interface or boundary.
The point where the incident ray strikes the interface is the point of incidence.
The angle between the incident ray and the normal to the interface is the angle of incidence \(i\).
The angle between the reflected ray and the normal is the angle of reflection \(r\).
A light ray strikes a reflective plane surface at an angle of \(56^\circ\) with the surface.
a) Find the angle of incidence.
b) Find the angle of reflection.
c) Find the angle between the reflected ray and the surface.
d) Find the angle between the incident and reflected rays.
a) \(i = 90^\circ - 56^\circ = 34^\circ\)
b) \(r = i = 34^\circ\) (law of reflection)
c) Let \(q\) be the angle between the reflected ray and the surface: \(q = 90^\circ - r = 56^\circ\)
d) Angle between incident and reflected rays: \(i + r = 68^\circ\)
A ray of light is reflected by two parallel mirrors (1) and (2) at points A and B. The ray makes an angle of \(25^\circ\) with the axis of the mirrors.
a) What is the angle of reflection at point A?
b) What is the angle of reflection at point B?
c) If the distance between the mirrors is \(d = 4 \text{ cm}\) and the length of the mirror system is \(L = 3 \text{ m}\), estimate the number of reflections.
d) If \(L\) and \(d\) are fixed, how can we reduce the number of reflections (and thus energy loss)?
a) \(i = 90^\circ - 25^\circ = 75^\circ \Rightarrow r = i = 75^\circ\)
b) Since mirrors are parallel, alternate interior angles are equal: \(i' = r = 75^\circ\)
c) \(\tan(75^\circ) = \frac{AC}{d} \Rightarrow AC = 4 \tan(75^\circ) \approx 15 \text{ cm}\)
Approximate number of reflections: \(N = \frac{L}{AC} = \frac{300 \text{ cm}}{15 \text{ cm}} = 20\)
d) Let the ray make an angle \(\alpha\) with the axis. Then \(i = 90^\circ - \alpha\), \(AC = d \cot(\alpha)\), and \(N = \frac{L}{d \cot(\alpha)}\). To decrease \(N\), increase \(\cot(\alpha)\) by decreasing \(\alpha\).
A light ray with angle of incidence \(i\) strikes mirror (1) at point A, then mirror (2) at point B, and then mirror (1) again at point C. The two mirrors form an angle \(\alpha\). Express the angles of incidence at points B and C in terms of \(i\) and \(\alpha\).
Let \(r = i\) (law of reflection). From geometry:
At point B: angle of incidence = \(i - \alpha\).
At point C: angle of incidence = \(i - 2\alpha\).
Thus, each reflection reduces the angle of incidence by \(\alpha\).
Find the angle \(\alpha\) between two mirrors such that the incident ray at A and the final reflected ray at B are parallel.
For the rays to be parallel, the sum of the angles of incidence and reflection at A and B must be supplementary:
\(i + r + i' + r' = 180^\circ\). Using \(r = i\) and \(r' = i'\), we get \(i + i' = 90^\circ\).
From the triangle formed by the normals and the mirror intersection: \(\alpha = i + i' = 90^\circ\).
Thus, the mirrors must be perpendicular (\(\alpha = 90^\circ\)).
