AP Physics B- Optics: Equations to be Remembered in Interference and Diffraction

If I have seen a little further it is by standing on the shoulders of Giants.

–Sir Isaac Newton

In Physical Optics the topics included are: (i) Interference and diffraction (ii) Dispersion of light and the electromagnetic spectrum. These topics carry 5% of the total points.

You must remember the following equations to make you strong in answering multiple choice questions involving interference and diffraction:

(1) Refractive index(n) of medium 2 with respect to medium 1 is given given by

n = n₂/ n₁ = λ₁/ λ₂ = v₁/v₂ where λ₁ and λ₂ are the wave lengths and v₁ and v₂ are the velocities of light in medium 1 and medium 2 respectively].

(2) The resultant displacement produced (at a particular point in a medium) by a number of waves is the vector sum of the displacements produced by each of the waves. For instance, if two waves arriving at a point P produce displacements y₁ and y₂, the resultant displacement at P is the vector sum of y₁ and y₂. If the interfering waves have amplitudes a₁ and a₂ and they arrive at the point P with a phase difference of φ, the resultant amplitude at P will be [a₁²+ a₂²+ 2a₁ a₂ cosφ]^1/2, in accordance with vector addition. If the interfering waves have intensities I₁ and I₂ the resultant intensity (I) at P will be given by

I = I₁ + I₂ + [2√(I₁I₂)] cosφ, since the intensity is directly proportional to the square of the amplitude.

If φ = 0, 2π, 4π, 6π,…..(an even multiple of π), the intensity is maximum (I_max) given by

I_max= I₁ + I₂ + [2√(I₁I₂)]

If φ = π, 3π, 5π, 7π,…..(an odd multiple of π),, the intensity is minimum (I_min) given by

I_min= I₁ + I₂ – [2√(I₁I₂)]

(3) Two sources of light (or any wave) are said to be coherent if they produce waves of the same frequency with a constant phase difference between them. (The condition of same frequency is implied in the condition of constant phase difference; yet it is usually specified when we define coherent sources). If two coherent sources produce waves of equal amplitude (a) and therefore equal intensity (I₀), then the resultant intensity (I) when they interfere is given by

I = I₀ + I₀ + [2√(I₀I₀)]cosφ = 2I₀(1+cosφ)

If the phase difference between the interfering waves (of equal intensity) is zero or an even multiple of π (in other words, if the path difference between the waves is zero or an integral multiple of the wave length λ), the intensity is maximum (I_max) given by

I_max = 2I₀(1+1) = 4I₀ …..(Path difference = nλ where n = 0,1,2,3,4,…..etc)

If the phase difference between the interfering waves (of equal intensity) is an odd multiple of π (in otherwords, if the path difference between the waves is an odd multiple of half wave length), the intensity (I_min) is zero given by

I_min = 2I₀(1–1) = 0……(Path difference = (2n+1)λ/2 where n = 0,1,2,3,4,….etc)

(4) In the interference fringe pattern produced by Young’s double slit, the distance of the n^th bright fringe (x_n) from the central (zero order) bright fringe is given by

x_n = nλD/d, where n = 0, ±1, ±2, ±3….etc.

where λ is the wave length of light, D is the distance between the screen (on which the fringes are formed) and the double slit and ‘d’ is the separation between the two slits S₁ and S₂ forming the double slit (fig.).

The distance of the n^th dark fringe (x_n) from the central (zero order) bright fringe is given by

x_n = (2n+ 1) λD/2d, where n = 0, ±1, ±2, ±3….etc. [We use ± sign for the integers to indicate the fringes on both sides of the central fringe]

The fringe width β is the distance between the centres of consecutive brigt fringes or dark fringes and is given by

β = x_n+1 –x_n

Therefore, β = λD/d

Angular fringe width is independent of the distance of the screen and is given by

β_θ = β/D = λ/d

(5) When light traveling through a rarer medium gets reflected from the surface of a denser medium, it suffers a phase change of π, which is equivalent to a path change (path difrference) of λ/2

(6) Colours in thin transparent films can be observed in reflected light as well as transmitted light. The condition for brightness in the reflected light is

2μtcosr = (2n+ 1)λ/2 where μ is the refractive index of the film, t is its thickness, r is the angle of refraction in the film and n = 0,1,2,3,4,….etc.

[A film of negligible thickness cannot satisfy the above condition and hence it cannot appear bright in reflected light. (It will appear dark)].

The condition for brightness in transmitted light is

2μtcosr = nλ where n = 0,1,2,3,4,….etc.

[A film of negligible thickness can satisfy the above condition (for n = 0) and hence it will appear bright in transmitted light].

You should note that Newton’s rings in reflected light are produced by the interference between the light waves reflected from the two surfaces of an air film contained in between a glass plate and a lens.

(7) In the diffraction pattern produced by a single slit, the angular separation between the first minima is 2λ/a where a is the width of the slit. The angular width of the central maximum is therefore 2λ/a.

The centre of the central maximum is at O (fig.). On either side of O, positions of minimum intensity are obtained when the angle θ = ±nλ/a where n = 1,2,3…etc.

Therefore, the minima are located at θ = ± λ/a, ± 2λ/a, ± 3λ/a…etc. (Fig.) Positions of

secondary maxima are obtained on either side of O corresponding to

θ = ± (2n+ 1)λ/2a where n = 1,2,3….etc.

Therefore, the secondary maxima are located at θ = ±3λ/2a, ±5λ/2a, ±7λ/2a….etc. The intensity of the secondary maxima goes on decreasing. [The first secondary maximum has intensity less than 5% of the intensity of the central maximum].

(8) In a plane diffraction grating, the n^th principal maximum is obtained at an angle θ given by

(a+b) sinθ = nλ where a is the width of the slit and b is the width of the opaque region between the slits. [a+b is called the grating element]

In the next post questions on wave optics will be discussed. Here is a question which you must be able to work out in no time if you have understood the phenomenon of interference.

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9

Work out the above question. I’ll be back shortly with the solution and of course, more solved questions.