Class 12 Physics Interference Notes

Class 12 Physics Interference Notes
UNIT 3
CLASS 12 PHYSICS • WAVE AND OPTICS

Interference

Chapter 11
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Coherent Source

Two sources are said to be coherent if they emit light wave of same wavelength, same amplitude and they are in same phase with each other having constant phase difference.

It is not possible two independent source are coherent but experimentally two virtual sources come from single source acts as coherent source.

Interference

When two waves of same frequency having constant phase difference travelling in same direction will meet, they produce resultant wave by superposition principle. Thus the modification in the intensity of light due to superposition of two or more wave is called interference.

Types of Interference

  1. Constructive interference
  2. Destructive interference

1. Constructive Interference

If two waves having same frequency and same phase overlap to each other, the amplitude of resultant wave is equal to the sum of amplitude of two wave that means a resultant wave of more intensity will be formed. Such interference is called constructive interference.
Constructive interference of two waves in same phase producing a larger resultant amplitude + = y₁ y₂ y = y₁ + y₂
Constructive interference
Path difference = nλ
where n = 0, 1, 2, 3 …

2. Destructive Interference

If two waves of same frequency and same amplitude travelling in same direction with opposite phase to each other, then resultant wave of zero amplitude is formed. Such interference is called destructive interference in which intensity of light will be minimum.
Destructive interference of two waves in opposite phase producing zero resultant amplitude + = y₁ y₂ y = y₁ + y₂
Destructive interference
Path difference = (n + 1/2)λ
where n = 0, 1, 2, 3 …

Conditions to Produce Interference

  1. Two source of light must be coherent.
  2. Light source used must be monochromatic.
  3. Two source of light must be closed to each other.
  4. The distance between coherent source of light and the screen should be large.
  5. Coherent source of light must be narrow.
  6. Coherent source of light must emit wave continuously.

Young’s Double Slit Experiment

Young’s double slit experiment showing source, two coherent slits, screen, central point and observation point S S₁ S₂ d P C N D x
Young’s double slit experiment

Let ‘S’ be the monochromatic source of light having wavelength ‘λ’. Also let S1 and S2 be the two slits which are equi-distance from source ‘S’ and act as coherent source.

Suppose ‘D’ be the distance of screen from coherent source and ‘d’ be the distance between two source as shown in figure above.

Also suppose ‘C’ be the centre of screen at which path difference is zero. Hence at point ‘C’ maximum intensity is observed. Also let ‘P’ be any point at distance ‘x’ from ‘C’.

Path Difference and Phase Difference

From the figure:

PN = x + d/2
PM = x − d/2

Also:

(S2P)2 − (S1P)2 = (PN)2 − (PM)2
(S2P − S1P)(S2P + S1P)
= (x + d/2)2 − (x − d/2)2
= [(x + d/2) + (x − d/2)] [(x + d/2) − (x − d/2)]
= (2x)(d)
= 2xd

If point P is very close to C:

S2P ≈ S1P ≈ D
S2P + S1P ≈ 2D

Therefore:

(S2P − S1P)2D = 2xd
Path difference = S2P − S1P = xd/D

Phase Difference

Phase difference = (2π/λ) × path difference
Phase difference = (2π/λ)(xd/D)

Bright Fringes

The point ‘P’ has maximum intensity if:

xd/D = nλ
x = nλD/d
xn = nλD/d

which gives the distance of nth bright fringe from centre.

For n = 1:
x1 = λD/d
For n = 2:
x2 = 2λD/d

Dark Fringes

For dark fringe, the path difference condition is:

xd/D = (n + 1/2)λ
xn = [(2n + 1)/2] λD/d
xn = (2n + 1)λD/(2d)

which gives the distance of nth dark fringe from centre.

For n = 0:
x0 = λD/(2d)
For n = 1:
x1 = 3λD/(2d)

Fringe Width

Fringe width is the separation between two successive bright fringes or two successive dark fringes.

For Bright Fringe

β = x2 − x1
= 2λD/d − λD/d
β = λD/d

For Dark Fringe

β = x1 − x0
= 3λD/(2d) − λD/(2d)
β = λD/d
Alternating bright and dark fringes with constant fringe width beta β/2 β Central bright
Alternating bright and dark interference fringes

Solved Numericals

Q.1 — Wavelength from Dark Fringe Separation

The separation between the consecutive dark fringes in a Young’s double slit experiment is 1 mm. The screen is placed at a distance of 2 m from the slits. If slit separation is 1 mm, what is the wavelength of light used in the experiment?

β = 1 mm = 1×10−3 m
d = 1 mm = 1×10−3 m
D = 2 m
λ = ?

Now:

β = λD/d
1×10−3 = λ×2/(1×10−3)
λ = 5×10−7 m

Q.2 — Slit Separation from Four Bright Fringes

In a Young’s double slit experiment, the separation of four bright fringes is 2.5 mm. The wavelength of light used is 6.2×10−7 m. If the distance from the slits to the screen is 80 cm, calculate the separation of two slits.

4β = 2.5 mm = 2.5×10−3 m
λ = 6.2×10−7 m
D = 0.8 m
d = ?

Now:

4λD/d = 2.5×10−3
4×6.2×10−7×0.8 / d = 2.5×10−3
d = 0.794 mm

Q.3 — Slit Separation and Angle for the Tenth Bright Fringe

In an experiment using Young’s slit, the distance between the centre of the interference pattern and the tenth bright fringe on either side is 3.44 cm. Distance between the slits and the screen is 2.0 m. If the wavelength of the light used is 5.89×10−7 m, determine the slit separation and the angle made by the central bright fringes at the slit.

D = 2 m
x10 = 3.44 cm = 3.44×10−2 m
λ = 5.89×10−7 m
d = ?

For the tenth bright fringe:

x10 = 10λD/d
3.44×10−2 = 10×5.89×10−7×2 / d
d = 3.42×10−4 m

Again, from the source working:

tanθ = x1/D = λ/d
= 5.89×10−7 / 3.42×10−4
= tan−1(1.72×10−3)
θ = 0.11 radian   [as written in the scan]
The final angular value written in the handwritten source is preserved exactly. It does not numerically match tan−1(1.72×10−3), so it has not been silently corrected.
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