Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: High

: Consider a slit of width ‘a’ illuminated by a parallel beam of monochromatic light of wavelength ‘λ’ The light is focused by lens L on the screen. Part of incident plane wave front is obstructed by the slit. Therefore, according to Huygens’s principle each and every point on the source act as a secondary source and emits secondary wavelets. Therefore, it is appropriate to replace slit as a string of point source which are coherent. These wavelets interfere constructively or destructively and produced interference pattern

Diffraction pattern due to single slit consists of wide central maxima flanked on either side by few alternate maxima and minima of falling intensity.

Most of the light travel straight and is focused at a point a P. All the wavelets are in phase up to slit AB. After the slit, they are again in phase when brought at ‘ P’ There wavelets reinforce each other and a bright band is obtained at ‘p’ is located at centre it is called as central maxima.

The wavelets which leave the slit at some angle ‘θ’ are focused at some other point on the screen. Let us consider point ‘Q’ on the screen.

To find resultant intensity at a point ‘Q’ . let us divide the slit AB into two equal parts AO and OB.

Waves ON and BM are in phase at slit, wave BM travel farther than ON by path difference of a/2 sinθ. If this path difference is λ/2. The two waves interfere destructively and produce darkness at This true for any two waves that originate at points separated by a/2 as the path difference between the from two such points will be λ/2. Therefore, it can be concluded that the waves from upper portion AO interfere destructively with the waves from lower portion OB. If a/2 sinθ=( λ)/2 a sin θ = λ condition for minima θ is of minimum intensity and it is called first order minima. We may also divide the slit into quarter, six and so on and by similar arguments. It can shown that minima occurs when A sin θ = λ, 2λ, 3λ, 4λ, …………. General condition for minima is a sinθ = nλ

To find resultant intensity at a point ‘ ..;;R’ after ‘Q’ let us divide the slit into three equal parts.

The optical path difference between the wavelets reaching ‘R’ from the corresponding points of first and second part is λ/2. This results the destructive interference at ‘R’ and the illumination at R is due to the third part of the slit. Hence there is some illumination at ‘R’ which will be much less than at ‘P’ since the net effect is only due to 1/3 rd of exposed part of slit hence ‘R’ to be bright
a sinθ = 3λ/2 same argument can be extended by dividing the slit into 5,7,9 ……. Parts Therefore maxima occurs when asinθ = 3λ/2, 5λ/2, 7λ/2 ………. General condition a sin θ = (2n ± 1) λ/2 a sin θ = m λ/2 ………………………………(2) The above two equation provides the conditions for minima and maxima but does not indicates anything about variation in intensity on screen. In order to determine intensity distribution, we follow graphical approach. Consider the slit divided into large number of narrow strips each of width dx.

One such slit shown below

The path difference between top and bottom wavelets diffracted from upper and lower edge of slit dx is dx sin θ

For path difference of λ phase difference is 2π

Therefore for path difference of dx sinθ , phase difference is

∆∅= (2π/λ) dx sinθ

Therefore, the total phase difference between the wavelets from and bottom of full slit of width ‘a’ is

∅= (2π/λ) a sinθ ……………….(3)

Each slit produces a disturbance of some amplitude this disturbance is represented by small vector known as phaser, the length of which gives the amplitude of disturbance for wavelets which travel straight has no phase difference between them. therefore, their phasers are laid end to end but no wavelets which leaves the slit at an angle θ has a phase difference of ∆φ with respect to the previous one. Therefore, phasers are represented as below

If the slit are large in no. their phasers describe a smooth arc AB of radius R

Φ = Total phase difference due to full slit.

$E_m$ = Maximum amplitude due to N slit ( arc AB )

$E_θ$ = Resultant amplitude ( chord AB )

Join the chord AB draw a perpendicular from C to AB In ∆ ADC Sin α = AD/AC

Sin α = $E_θ$ / 2R

$E_θ$ = 2Rsin α ……………………(4)

Angle φ = Arc AB / Radius

φ =$E_m$ / R

R=$E_m$ / ∅

$E_θ$ = (2$E_m$ sinα) / ∅

$E_θ$ = ($E_m$ sinα) /(∅⁄2)

$E_θ$ = ($E_m$ sinα) / ∝ …………………………(5)

Intensity is square of amplitude

$E_θ^2$ = ($E_m^2$ $sin^2$ α)/α^2

$I_θ$ = $I_m$ ($sin^2$ α)/ $α^2$ …………………(6)

To find maxima and minima. differentiate eq. (6) and set it to zero dI/dα=0 $I_m$ [(2 sinα.cosα)/ $α^2$ - ($sin^2$ α)/ $α^3$ ]=0

($I_m$ sinα/ $α^3$ ) [α cosα-sinα]=0

Sin α = 0 α.cos α – sin α = 0

Sin α = 0 α = tan α

Sin α = 0 determines the condition for minima value of α which satisfies the conditions are

α= $ I_m$ π

∅/2= ± $I_m$ π

(π/λ) a sinθ= ±mπ

a sinθ= ±mλ ………………….(7)

This is the condition for minima.