Diffraction of Light Formulas

1. Diffraction

Bending of waves around the edges of an obstacle or deviation from j rectilinear propagation is called diffraction. For the observation of diffraction the size of the obstacle ΓÇÿaΓÇÖ must be of the order of wavelength ╬╗, of the waves i.e., a ~ ╬╗.

2. Types of diffraction

(a) Fresnel: In which source and screen are at finite distance wavefronts are curved.
Fresenl’s diffraction:

(b) Fraunhofer:
In which source and screen are effectively at infinite distances i.e. wave fronts are plane.
FraunhoferΓÇÖs diffraction:

3. FresnelΓÇÖs half period zones

The wave front can be divided into zones such that waves reaching a given point from successive zones differ in phase by π, path by \(\frac{\lambda}{2}\) and time by \(\frac{T}{2}\).

For a plane wave front- Radius of n^th zone r_n = \(\sqrt{n b \lambda}\), r_n ∝ \(\sqrt{n}\)
Area of n^th zone

Mean distance from the given point
d_n = b + (2n – 1)\(\frac{\lambda}{4}\)
If source is at a distance ΓÇÿaΓÇÖ from obstacle i.e, wave front is spherical then-
r_n = \(\left[n\left(\frac{a b^{*}}{a+b}\right) \lambda\right]^{1 / 2}\)

4. Amplitude of waves reaching a point

Amplitude R_n ∝ \(\frac{\mathrm{A}_{\mathrm{n}}\left(1+\cos \theta_{\mathrm{n}}\right)}{\mathrm{d}_{\mathrm{n}}}\)
As n increases d_n and ╬╕_n increase so R_n decreases.
R₁ > R₂ > R₃………,
R₂ = \(\frac{\mathrm{R}_{1}+\mathrm{R}_{3}}{2}\), R₄ = \(\frac{\mathrm{R}_{3}+\mathrm{R}_{5}}{2}\), ……….
and \(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{R}_{3}}{\mathrm{R}_{2}}=\ldots \ldots . .=\frac{\mathrm{R}_{\mathrm{n}}}{\mathrm{R}_{\mathrm{n}-1}}\) = constant

5. Resultant amplitude

R = R₁ – R₂ + R₃ – R₄ …….. + (-1)ⁿ R_n
R = \(\frac{\mathrm{R}_{1}}{2}+\frac{\mathrm{R}_{\mathrm{n}}}{2}\) if n is odd.
R = \(\frac{R_{1}}{2}-\frac{R_{n}}{2}\) if n is even.
For a large n, Rⁿ is negligible.
∴ R = \(\frac{R_{1}}{2}\)
Intensity Γê¥ (Amplitude)²,
I = KR²
Resultant intensity due whole wavefront
I = \(\frac{\mathrm{KR}_{1}^{2}}{4}=\frac{\mathrm{I}_{1}}{4}\)

6. Diffraction due to an opaque disc

The disc blocks certain number of half period zones. Numher of zones blocked

n = 1, I’ = \(\frac{\mathrm{KR}_{2}^{2}}{4}\)
n = 2. I” = \(\frac{\mathrm{KR}_{3}^{2}}{4}\) ………..
I > I’ > I”. …………..
Centre is always bright but intensity decreases as number of zone obstructed increases.

7. Diffraction due to a circular aperture

The aperture allows waves from certain number of half period zones only.

If n = 1, only one zone is exposed I₁ = KR₁² = 4 times the intensity due to whole wavefront.
If n = 2, I₂ = K(R₁ – R₂)² Γëê 0, centre is dark.
n = odd number, the centre is bright and when n = even number centre is dark. Radius of first dark ring (AiryΓÇÖs ring) around central bright patch
x₁ = \(\frac{b \lambda}{2 r}\)
A lens can be treated as a circular aperture with 2r = diameter of lens D, b = focal length of lens f then
x₁ = \(\frac{f \lambda}{D}\)
More accurate analysis gives

8. Zone plate

A transparent plate on which circles of radii proportional to square root of natural numbers are drawn and alternate zones are made opaque.
Positive zone plate-Odd zones are transparent and even are opaque.
Negative zone plate-Odd zones are opaque and even transparent.
It behaves as a convex lens having multiple focii.
Positive zone plate:

Negative zone plate:


f_p = \(\frac{r^{2}}{(2 p-1) \lambda}\), r is radius of first circle on plate.
= \(\frac{r_{n}^{2}}{(2 p-1) n \lambda}\), r_n is radius of nth circle on plate.
f₁ = \(\frac{r^{2}}{\lambda}\) (Principal focal length), f₂ = \(\frac{r^{2}}{3 \lambda}\), ………..
If ΓÇÿaΓÇÖ is distance of object and ΓÇÿbΓÇÖ is distance of screen from plate the

9. Fraunhofer diffraction due to a single slit

On both sides of central maximum diffraction fringes of unequal thickness are obtained. Angular position of minima are given by
a sin ╬╕_n = n╬╗ (n = 1, 2, …………)
Γê┤ ╬╕₁ Γëê \(\frac{\lambda}{a}\)
Angular width of central maximum = 2╬╕₁ = \(\frac{2 \lambda}{\mathrm{a}}\)
If focused by a lens of focal length f, the linear width = \(\frac{2 \mathrm{f} \lambda}{\mathrm{a}}\)
Angular positions of maxima are given by
a sin ╬╕_n = (2n + 1) \(\frac{\lambda}{2}\)

10. Resolving power

The smallest separation between two point objects whose images can be NOTES j seen separate i.e, resolved, is called limit of resolution.
Resolving power = \(\frac{1}{\text { Limit of resolution }}\)

11. ReyleighΓÇÖs criterion

Two objects are just resolved if in the diffraction pattern central maximum of first lies at first minimum of the other and vice-versa.

12. Telescope

Limit of resolution of telescope
╬▒ = \(\frac{1.22 \lambda}{a}\)
∴ Resolving power = \(\frac{1}{\alpha}=\frac{a}{1.22 \lambda}\)
ΓÇÿaΓÇÖ is aperture of objective of telescope.

13. Microscope

Limit of resolution d = \(\frac{1.22 \lambda}{2 \sin \theta}=\frac{0.61 \lambda}{\sin \theta}\)
2╬╕ is angle subtended by the objective at the position of objects. If objects are immersed in a medium of refractive index p then-
Limit of resolution d = \(\frac{0.61 \lambda}{\mu \sin \theta}\)
∴ Resolving power = \(\frac{1}{d}=\frac{\mu \sin \theta}{0.61 \lambda}\)
┬╡ sin ╬╕ is called numerical aperture.

14. Prism

Resolving power = \(\frac{\lambda}{d \lambda}=t \frac{d \mu}{d \lambda}\), (t is width of base of prism)

15. Human eye

Limit of resolution = 1 minute of arc = \(\left(\frac{1}{60}\right)=\frac{1}{60} \frac{\pi}{180}\) radian.

Get a Complete List of Physics Formulas on Physicscalc.Com to get help on various concepts in no time.