Subject: Applied Physics 2

Topic: Interference And Diffraction Of Light

Difficulty: High

A diffraction grating consists of a closely spaced aperture of width ‘a’ separated by opaque interval of width ‘b’. AB, CD, EF …. are apertures and BC, DE, FG are opaque parts. Consider point A and C on grating. These are called corresponding points. The distance between any such pair of points equals to (a+b) and is called grating element or grating constant. If there are N apertures and N opaque interval in 1 m then N(a+b) = 1

$\therefore a+b = 1/ N$

Grating element is equal to the reciprocal of number of lines per cm on grating.

Let a train of plane waves be incident normally on grating. Considering light rays passing through the grating straight will be conveyed at ‘P’ As the wavelets through the various slits reach the point ‘p’ after covering equal distance.‘P’ is called as zero order principle maxima.

Let us consider the light leaving the various slits at an angle θ with that of incident beam. From point A draw the normal. There is the path difference between the rays starting from various slit and reaching θ

Suppose θ is such that CL = λ , EM = 2λ, GN = 3λ and so on.

The waves from all these elements are in phase at grating and are also in phase along the line AR and reach Q in phase.

Therefore, the reinforce each other and produced first order principle maxima at Q.

If θ such that CL = 2λ then again waves will be in phase and produce second order maxima.

In ∆ACL sin⁡θ= λ / (a+b)

(a+b) sin θ = λ

In general (a+b) sin θ =n λ

Each slit in the grating produces its own diffraction pattern.since the slits are very large in number.only few maxima are seen whereas other maximas and minimas are suppressed.

To find resultant intensity all the secondary wave in each can be replaced by a single slit of amplitude ($E_m$ sinα) / α starting from the midpoint of the slit and traveling at an angle θ with normal.

Let the path difference between the waves starting from midpoint of slit equal to λ

For path difference λ, phase difference 2π

For path difference (a+b) sin θ, phase difference ∆∅= (2π/λ) (a+b)sinθ

∆∅=2β β= ( π/λ) (a+b)sinθ ………………….(1)

To find the resultant amplitude, we have to find the resultant of N vibration of amplitude ($E_m$ sinα) / α and the phase difference is 2β. And it is found out by vector addition method

$E_θ$ = ($E_m$ sinα / α ) sinNβ / sinβ …………………………………(2)

We know that I α $E^2$

$E_θ^2$ = [$E_m^2$ ($sin^2$ α ) / $α^2$ ] [($sin^2$ Nβ ) / ($sin^2$ β)]

$I_θ$ = $I_m$ [ ($sin^2$ α ) / $α^2$ ] [ ($sin^2$ Nβ ) / ($sin^2$ β) ] …………………………..(3)

The first factor ($sin^2$ α) / $α^2$ in equation (3) gives the intensity distribution due to single slit and second factor

($sin^2$ Nβ) / ($sin^2$ β) gives the intensity pattern due to N slits

For $I_θ$ to be maximum sin β = 0

But sin β = 0 sin Nβ = 0

Using L hospital rule

lim┬(β→±mπ)⁡〖d(sinNβ)/d(sinβ) 〗

=N (cosNβ )/cosβ

=N (cosNmπ )/(cosmπ )

=N

Therefore ($sin^2$ Nβ ) / ($sin^2$ β) = $N^2$

So equation (3) becomes

$I_θ$ =$N^2$ $I_m$ ($sin^2$ α) / $α^2$

Condition for maxima β = ±mπ

[ π/λ (a+b)sinθ] = ±mπ

(a+b)sinθ= ±mλ m = 1,2,3,4…………………

For minima Nβ = ±mπ N ( π)/λ (a+b)sinθ= ±mπ N(a+b)sinθ= ±mλ m can be taken all the values except 0,N,2N,……………nN Because for all the values condition for maxima will be satisfied therefore m can be (nN±1) values.

Determination of Wavelength of Light using Diffraction Grating

The diffraction grating is often used in the laboratories for the determination of wavelength of light. The grating spectrum of the given source of monochromatic light is obtained by using a spectrometer. The arrangement is as shown in Figure shown below The spectrometer is first adjusted for parallel rays. The grating is then placed on the prism table and adjusted for normal incidence. In the same direction as that of the incident light, the direct image of the slit or the zero-order spectrum can be seen in the telescope. On either side of this direct image a symmetrical diffraction pattern consisting of different orders can be seen. The angle of diffraction θ for a particular order m of the spectrum is measured.

The numbers of lines per inch of grating are written over it by the manufacturers.

Thus using the equation, a + b) sin θ = mλ

The unknown wavelength λ can be calculated by putting the values of the grating element (a + b), the order m and the angle of diffraction θ .