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## Wednesday, January 9, 2008

### 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 (Imax) given by
Imax= I1 + I2 + [2√(I1I2)]
If φ = π, 3π, 5π, 7π,…..(an odd multiple of π),, the intensity is minimum (Imin) given by
Imin= 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 (Imax) given by
Imax = 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 (Imin) is zero given by
Imin = 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 nth bright fringe (xn) from the central (zero order) bright fringe is given by
xn = 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 nth dark fringe (xn) from the central (zero order) bright fringe is given by
xn = (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 = xn+1 xn
Therefore, β = λD/d
Angular fringe width is independent of the distance of the screen and is given by
βθ = 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 = 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 nth 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.