# Interference of Light Formulas

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## Cheat Sheet on Interference of Light

1. Interference

By superposition of waves of same frequency and constant phase difference (coherent waves) redistribution of energy takes place so intensity is maximum at certain points and minimum at others. This phenomenon is called interference.

2. Principle of superposition

y = y1 + y2 + …….. + yn
For two waves y1 = a sin ωt
and y2 = b sin (ωt + Φ)
Resultant Amplitude A = (a2 + b2 + 2ab cos Φ)1/2
tan θ = $$\frac{b \sin \phi}{a+b \cos \phi}$$
(θ → phase difference between y1 & y2)
Intensity I ∝ A2
∴ I = I1 + I2 + 2$$\sqrt{I_{1} I_{2}}$$ cosΦ
If a = b = a0, then I1 = I2 = I0
∴ A = 2a0 cos $$\left(\frac{\phi}{2}\right)$$
and I = 4I0 cos2$$\left(\frac{\phi}{2}\right)$$

3. Constructive interference (a) Condition – Phase difference Φ = 0, 2π,…. 2nπ
path difference x = 0, λ, 2λ ……… nλ.
A = Amax = (a + b),
I = Imax = $$\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$$ = (a + b)2

(b) If a = b = a0, I1 = I2 = I0
Amax = 2a0 and Imax = 4I0

4. Destructive interference

(a) Condition – Phase difference
Φ = π, 3π, = (2n+ 1)π
path difference
x = $$\frac{\lambda}{2}, \frac{3 \lambda}{2}, \ldots . .=(2 n+1) \frac{\lambda}{2}$$
A = Amin = (a – b)
I = Imin = $$\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}$$ = (a – b)2

(b) If a = b = a0, I1 = I2 = I0
Amin = 0, Imin = 0 5. Coherent source

Two identical source which have

• Same amplitude.
• Same frequency.
• Constant phase difference, are called coherent source.

6. Methods of obtaining coherent sources

• By division of wavefront and
• By division of amplitude

7. Conditions for sustained interference

• Coherent waves
• Amplitudes nearly equal for better contrast and
• Same polaristion

8. Formation of fringes and fringe width (young’s double slit exp.) (a) Fringe width
β = $$\frac{\lambda \mathrm{D}}{\mathrm{d}}$$ (same for bright and dark fringes)

(b) Position of nth bright fringe
xn = n $$\frac{\lambda \mathrm{D}}{\mathrm{d}}$$

(c) Position of nth dark fringe
xn‘ = (2n – 1) $$\frac{\lambda \mathrm{D}}{\mathrm{d}}$$

(d) Angular width of fringe
θ = $$\frac{\beta}{D}=\frac{\lambda}{d}$$

(e) Fringe visibility or contrast
Q = $$\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{2 \sqrt{I_{1} I_{2}}}{I_{1}+I_{2}}$$

(f) Shape of fringes in space is hyperbolic and the intercept on screen gives straight line fringes.

(g) If whole exp. is done in water then p reduces, (β ∝ $$\frac{1}{\mu}$$)

9. Effect of thin sheet placed in the path of one of the waves

Thin sheet introduces an extra path difference = (µ – 1) t.
If central fringe is shifted to position of nth bright fringe then
(µ – 1) t = nλ
If central fringe is shifted to position of n,h dark fringe then
(µ – 1) t =(2n – 1)$$\frac{\lambda}{2}$$

10. Fresnel’s biprism  (a) Coherent sources produced by refraction from the two parts of biprism

(b) d = 2a(µ – 1)α
D = (a + b), µ is refractive index of prism of angle α.

(c) d = $$\sqrt{\mathrm{d}_{1} \mathrm{d}_{2}}$$ , d1 and d2 are distances between images of the coherent sources in the two conjugate positions of lens.

(d) β = $$\frac{\lambda \mathrm{D}}{\mathrm{d}}$$, λ = $$\frac{\beta d}{D}$$

(e) If whole apparatus is placed in a medium of refractive index aµm then both d and λ are affected-
βm = $$\frac{\lambda_{m} D}{d_{m}}$$
λm = $$\frac{\lambda_{\mathrm{air}}} { }_{\mathrm{a}} \mu_{\mathrm{m}}}$$
dm = 2a(mµg – 1)α

(f) If whole exp. is done in water the β increases.

11. Interference by thin films (a) Interference occurs between the waves reflected from the upper and lower surfaces of the film

(b) Effective path difference introduced in reflected waves
x = 2µt cos r + $$\frac{\lambda}{2}$$ = nλ (max)
= (2n – 1) $$\frac{\lambda}{2}$$ (min)
∴ 2µt cos r = (2n – 1) $$\frac{\lambda}{2}$$ (max)
= nλ (min)

(c) If t ≈ 0, very thin film, x ≈ 0, film appears dark in reflected light.

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