Light is a form of energy, which induces the sensation of vision in our eyes and makes us able to see various things present in our surroundings. The light ray may be objects self-light or reflected light.
Fig: Wave nature of lightIt travels in straight line in form of particles and waves. With the help of light we see all colours of nature.
Our eyes are mostly sensitive for yellow colour and least sensitive for violet and red colour. Due to this reason commercial vehicle's are painted with yellow colour, sodium lamps are used in road lights.
Properties of light:
Light energy propagates (travels) via two processes.
(i) The particles of the medium carry energy from one point of the medium to another.
(ii) The particles transmit energy to the neighbouring particles and in this way energy propagates in the form of a wave.
(iii) It propagates in straight line.
(iv) It's velocity in vacuum is maximum whose value is 3 × 108 m/sec. (297489978 m/s)
(v) Light does not need any material medium to travel, that is it can travel through a vacuum.
(vi) It exhibits the phenomena of reflection, refraction, interference, diffraction, polarization and double refraction.
Ray of light:
A straight line which shows the direction moment of light is called ray of light.
Beam of light:
A bunch of light rays or bundle of rays at a point is called beam of light.
How do we see?
When a light ray is falling (strike) on the surface of any object it reflects and reaches to our eyes. Due to this our eyes feel a sensation then we see the object.
Reflection of light:
When rays of light falls on any object it returns back in the same medium from the surface this phenomenon is called reflection of light. Due to reflection of light we can see all the nature.
The ray of light which falls on a polished surface (or a mirror) is called the incident ray of light.
The ray of light which gets reflected from a polished surface (or a mirror) is called the reflected ray of light.
The normal is a line at right angle to the reflecting surface.
Fig: Reflection of light
Laws of reflection:
(i) The incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in the same plane.
(ii) The angle of incidence (i) is always equal to the angle of reflection (∠r) i.e. ∠i = ∠r
(iii) When a ray of light falls on a mirror normally or at right angle it gets reflected back along the same path.
Depending on the nature of the reflecting surface there are two types of reflection:
(i) Regular (specular) reflection
(ii) Irregular (diffused) reflection
When parallel light rays fall on smooth plane surface like mirror, all rays of light are reflected parallel along a definite direction. Then this kind of reflection is called regular reflection.
Fig: Regular reflection
Irregular reflection (Diffused reflection):
Fig: Irregular Reflection
When parallel light rays fall on a rough surface all the rays of light are reflected in all possible (Different) directions this is called diffused or irregular reflection.
Example 1: A light ray strikes a reflective plane surface at an angle of 56° with the surface.
a) Find the angle of incidence.
b) Find the angle of reflection.
c) Find the angle made by the reflected ray and the surface.
d) Find the angle made by the incident and reflected rays.
We use the diagram shown below to answer the questions.
a) angle of incidence: i = 90 - 56 = 34°
b) angle of reflection r = i = 34 ° (by the law of reflection)
c) q = 90 - r = 90 - 34 = 56°
d) i + r = 34 + 34 = 68°
Example 2: 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° with the axis of the two mirrors.
a) What is the angle of reflection at the point of incidence A?
b) What is the angle of reflection at the point of incidence B?
c) If the distance between the two mirrors id d = 4 cm and the length L of the two mirror system is 3 meters, approximately how many reflections take place between the two mirrors?
d) In a real system, at each reflection, there are losses of the light energy travelling between the two mirrors. If L and d are fixed, what can be done to decrease the number of reflections on the mirrors and hence the energy lost by reflection?
Let's use the diagram shown below to answer the questions. n1 and n2 are normal to the reflecting mirrors and are therefore perpendicular to the axis of the two mirror system.
a) i = 90 - 25 = 75 °
b) The two mirrors are parallel and AB intersect both of them, hence angle r and i' are alternate interior and therefore equal in size.
Also r = i (law of reflection).
Hence i' = r = i = 75 °
c) If the distance between the two mirrors is = 4 cm, AC can be calculated as follows
tan(75°) = AC / d = AC / 4
AC = 4 tan(75°) = 15 cm
The number of reflections N that will take place between the two mirrors may be approximated as follows.
N = L / AC = 3 m / 15 cm = 20 cm
d) Let the angle made by the light ray and the axis of the two mirror system be α instead of 25 °.
The angle of incidence i = 90 - α = r = i'
tan(i') = tan( 90 - α) = cot (α) = AC / d
AC = d cot(α)
Number of reflections = N = L / AC = L / d cot(α)
If L and d are fixed, the only way to decrease N is to increase cot(α) and hence decrease α.
Example 3: A light ray at an angle of incidence i first strikes mirror (1) at point A, then mirror (2) at point B and then mirror (1) again at point C. The two mirrors make an angle α. Express the angles of incidence at point B and point C in terms of the angle of incidence i and angle α?
We first complete the given diagram drawing the various rays at points of incidence. Diagram shown below.
Let AD and CE be the normals to mirror (1) at A and C respectively.
r = i (law of reflection)
In triangle OAD, we can write
α + 90 + r' = 180 °
Simplify to get: r' = 90 - α
In triangle ADB, we have
r + r' + r" = 180 °
r" = 180 - r' - r
Substitute r' by 90 - α and r by i in r" to obtain
r" = 90 + α - i
The angle of incidence at B is given by
90 - r" = 90 - (90 + α - i) = i - α
In triangle CBE, we have
∠CBE = r" (incident ray and reflected ray make angles with surface of reflection of equal size)
In triangle OCB, we have
α + 90 + i' + ∠CBE = 180 °
Substitute ∠CBE by r"
α + 90 + i' + r" = 180 °
Substitute r" by 90 + α - i to obtain
α + 90 + i' + 90 + α - i = 180 °
Simplify to obtain the angle of incidence i' at C
i' = i - 2 α
At each reflection, the angle of incidence decreases by an amount equal to the size of angle α.
Example 4: Find angle α made by the system of the two mirrors shown in the figure below so that the incident ray at A and the reflected ray at B are parallel.
Solution: We first complete the given diagram with the angles of incidence and reflection as shown below and also labeling the incident and reflected rays.
For the incident ray at A and the reflected ray at B to be parallel, angles i + r and i' + r' have to be supplementary.(geometry: parallel lines cut by a transversal).
i + r + i'+ r' = 180 °
by law of reflection : r = i and r' = i'
Substitute to obtain
i + i + i' + i' = 180 °
i + i' = 90
In triangle AOB, we have
α + (90 - r) + (90 - i') = 180 °
α = r + i' = i + i' = 90 °
If α = 90 °, the incident ray at A and the reflected ray at B are parallel.