Position of Maxima & Minima | Oscillations, Waves & Optics - Physics PDF Download

Condition for Principle Maxima

When the value of N is very large, one obtains intense maxima when sinβ = 0 . This condition arises when

β = ±nπ where n = 0,1,2,3.....

so sin Nβ = 0

Thus  (indeterminate form), thus

Thus intensity

Such maxima are most intense and are known as principal maxima. Physically, at these maxima the fields produced by each of the slits are in phase, and therefore, they add and

the resultant field is N times the field produced by each of the slits.

They are obtained in the direction given by β = ±nπ

(b + e)sinθ = ±nλ where n = 0,1,2...

This is the condition of maxima where for n = 0 , we get the zero order principle maximum. For n = ±1,+2 + 3..... we obtain the first, second, third- order principal maxima respectively. The ± sign show that there are two principal maxima for each order

lying on either side of the zero order maximum.

Condition for Minima

The intensity is zero when sin Nβ = 0 but sinβ ≠ 0 then = 0 

( sinβ = 0 gives maximum intensity). Thus minimum are obtained in the directions given by

sin Nβ = 0 ⇒ Nβ = ±m′π ⇒

Thus the condition of minimum intensity is

N (b + e)sinθ = ±m′λ

where m′ takes all integral values except 0, N,2N.... nN . Because these value of m′ makes sinβ = 0 , which gives principal maxima.

It is clear from above that m′ = 0 gives the principal maximum and m′ = 1, 2,3.. (N −1) give minima and then m′ = N give again a principal maximum. Thus there are (N −1) minima between two consecutive principal maxima.

Condition for Secondary Maxima

As there are (N −1)minima between two consecutive principal maxima, there must be (N − 2) other maxima between two principal maxima these are called secondary maxima.

Their positions are obtained by differentiating the formula of intensity I with respect to

β and equating it equal to zero. Thus

 = 0

N cos nβ sinβ = sin Nβ cosβ ⇒ tan Nβ = N tanβ                              ----(1)

To find the value of  under the condition (1), we make use of the triangle shown

in figure (1), this gives

This gives

This shows that the intensity of the secondary maxima is proportional to, where the intensity of the principal maxima is proportional to N² so,

Hence greater the value of N , the weaker are secondary maxima. In an actual grating, N

is very large. Hence these secondary maxima are not visible in the grating spectrum.

Condition for absent spectra

A particular principal maximum may be absent if it corresponds to the angle which also determines the minimum of the single-slit diffraction pattern.

Principal maxima in the grating spectrum are obtained in the direction given by
(b + e)sinθ = nλ                                                          -----(1)

where n is the order of maximum.

The minima in a single slits pattern are obtained in the direction given by

bsinθ = mλ m =1,2,3...                                                -----(2)

If both (1) and (2) are satisfied simultaneously, a particular maximum of order n will be

missing in the grating spectrum.

this is the condition for the spectrum of the order n to be absent.

If e = b then n = 2m = 2,4,6... i.e. 2^nd, 4^th, 6^th, order spectra will be absent

If e = 2b then n = 3m = 3,6,9... i.e. 3^rd, 6^th, 9^th…… order spectra will be absent

Example 4: Monochromatic light beam a helium-neon laser (λ = 632.8 nm) is incident normally on a diffraction grating containing 6000 grooves per centimeter. Find the angles at which the 1st and 2nd order maxima are observed. What if we look for the 3^rd order maxima? Do we find it?

First we must calculate the slit separation, which is equal to the inverse of the

number of grooves per centimeter.

d = 1/6000cm = 1.667 x 10^-4 cm = 1667 nm

For the 1^st order maximum (n =1) , we obtain

 = 0.3796 ⇒ θ₁ = 22.31^o

For the 2^nd order maximum (n = 2) , we find

 = 0.7592 ⇒ θ₂ = 49.39^o

For 3^rd order maximum (n = 3) , we find

= 1.139

because sinθ cannot exceed unity, this does not represent a realistic solution. Hence only zeroth 1^st & 2^nd order maxima are observed for this situation

Rayleigh Criterion of Resolution and Resolving Power

The resolving power of an optical instrument represents its ability to produce distinctly

separate spectral lines of light having two or more close wavelengths.

Rayleigh’s Criterion of Resolution

Lord Rayleigh proposed the following criterion for resolution which has been universally adopted. “Two spectral lines of equal intensities are just resolved by an optical instrument when the principal maximum of the diffraction pattern due to one falls on the first minimum of the diffraction pattern of the other.

Resolving Power of a Grating

The resolving power of a grating represents its ability to form separate spectral lines for

wavelengths very close together. It is defined by R = λ/Δλ where Δλ is the separation of

the two wavelengths which the grating can just resolve; the smaller the value ofΔλ , larger the resolving power.

Let a parallel beam of light of two wavelengths of λ and λ + dλ be incident normally on the grating. If the n^th principal maximum of λ is formed in the direction n θ_n , we have
(b + e)sinθ_n = nλ                                                                                    (1)

where (b + e) is the grating element. Let the first minimum adjacent to the

nth maximum be obtained in the direction (θ_n + dθ_n )

The equation for the minima is
N (b + e)sinθ = m′λ                                                                               (2)
here θ = θ_n + dθ_n 

where N is the total number of rulings on the grating and m′ takes all integral values except 0, N,2N,....n N because these values of m′ give 0^th ,1^st ,2^nd ,....n^th principal maximum respectively.

Clearly, the first minimum adjacent to the principal maximum in the direction of θ increasing will be obtained for m′ = (nN +1) . Therefore, if this minimum is obtained in

the direction θ_n + dθ_n , we have from equation (2)

N (b + e)sin(θ_n + dθ_n) = (nN +1)

N (b + e)sin(θ_n + dθ_n) =                                                       (3)

By Rayleigh’s criterion, the wavelengths λ and (λ + dλ ) are just resolved by the grating

when the nth maximum of (λ + dλ ) is also obtained in the direction θ_n + dθ_n . Then, we

have from equation (1)

(b + e)sin(θ_n + dθ_n) = n λ + dλ                                                             (4)

Comparing equation (3) and (4), we get

n.Nλ +λ = Nnλ + +Nndλ

λ = Nndλ

λ/dλ = nN

But λ/dλ is the resolving power R of the grating.

Therefore, R = λ/dλ = nN

The above expression may be written as

R = nN =

But N (b + e) is the total width of the grating. Hence at a particular angle of diffraction θ_n, the Resolving Power is directly proportional to the total width of the ruled space on the grating.

Difference between Dispersive Power and Resolving Power

The dispersive power of a diffraction grating gives us an idea of the angular separation between the lines of a spectrum produced by a grating, it is measured by where dθ

is the angular separation between two spectral lines whose wavelengths differ by dλ .

The value of Dispersive Power is given by

Thus, higher is the order n of the spectrum or closer are the rulings on the grating (i.e. smaller the value of (b + e) greater is the dispersive power.

The resolving power of the grating, on the other hand, expresses the degree of closeness

which the spectral lines can have and yet be distinguished as two. It is measured by λ/dλ,

where dλ is the smallest wavelength λ . The value of Resolving Power is given by

λ/dλ = nN

where, N is the total number of rulings on the grating and n is order of diffraction. Thus

greater is the width of the ruled surface, higher is the resolving power. The higher

resolving power results in sharp maxima.

Resolving power is greater in case (a) than case (b).

Example 5: When a gaseous element is raised to a very high temperature, the atoms emit radiation having discrete wavelengths. Tow strong components in the atomic spectrum have wavelengths of 5890 A⁰ & 5896 A⁰ .
(a) What resolving power must a grating have if these wavelengths are to be distinguished?
(b) To resolve these lines in the second-order spectrum, how many slits of the grating must be illuminated?

(a) Resolving power is defined as

 where λ = = 5890 + 5896/2 = 5893 A⁰

Δλ = λ −λ = 5896 − 5890 = 6 A⁰

∴ R =  = 982 ⇒ R = 982

(b) Resolving power also defined as R = nN where n is the order of spectrum and N is number of slits

N = R/n = 982/2 = 491 slits

Example 6: A beam of light is incident normally on a diffraction grating of width 1 cm. It is found that at 30^o, the n^th order diffraction maximum for λ₁ = 600 Ǻ is super-imposed on the (n+1)^th order for λ₂ = 500 Ǻ. How many lines per cm does the grating have? Find out whether the first order spectrum from such a grating can be used to resolve the wavelengths λ₃ = 5800 Ǻ and λ₄ = 5802 Ǻ?

Condition for diffraction maxima is

(b + e)sinθ = nλ

For given two wavelength the maxima condition can be written as

(b + e)sin 30^o = nλ₁ and (b + e) sin 30^o = (n +1)λ₂

_{Taking ration of the two, we get}

The grating element is

(b + e) sin 30^o = 5×500Α⁰ ⇒ b + e = 5×10⁻⁵cm

Number of grating is

N = Grating width/Grating element = W/(b + e) = 1/5 x 10^-5 = 20000

Resolving power of this Grating is

R = λ/dλ = nN

For 1^st order spectrum, the resolving power of the given grating is

R = nN = 20000

Whereas to resolve the wavelengths λ₃ = 5800 Ǻ and λ₄ = 5802 Ǻ the required resolving power is

Resolving power of the given grating (20000) is more than required(2900) . Thus, given grating will resolve the wavelengths λ3 = 5800 A⁰ and λ₄ = 5802 A⁰.