IEMDaily - Video Lecture Notes

Example

Resultant Intensity of two superimposing sound waves with same frequency

Example:

S_{1}

and

S_{2}

are two coherent sources of sound of frequency 110Hz each. They have no initial phase difference. The intensity at a
point P due to

S_{1}

I_{0}

and due to

S_{2}

4 I_{0}

. If the velocity of sound is

330 m / s

then what is the resultant intensity at P?

Solution:
Wavelength

λ = \frac{v}{f} = \frac{330}{110} = 3 m

as velocity and frequency are given.
The expression for intensity of two waves after interference is

I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ

.....(1) where

ϕ

is the phase difference due to path difference (

S_{2} P - S_{1} P

)
As seen from the figure,

S_{2} P = \sqrt{3^{2} + 4^{2}} = 5

S_{1} P = 4

ϕ = \frac{2 π}{λ} (S_{2} P - S_{1} P) = \frac{2 π}{λ} \times (5 - 4) = \frac{2 π}{3} \times 1 = \frac{2 π}{3}

Substituting

ϕ

I_{1} = I_{0}

I_{2} = 4 I_{0}

in (1)

I = I_{0} + 4 I_{0} + 2 \sqrt{I_{0} \times 4 I_{0}} \cos \frac{2 π}{3} = 5 I_{0} + 4 I_{0} \times (- \frac{1}{2}) = 5 I_{0} - 2 I_{0} = 3 I_{0}

Example

Use vector method for addition of two superimposing sound waves of same frequency

Example: Two wires having different densities are joined at

x = 0.

An incident wave

y = A_{0} s i n (ω t - k_{1} x)

travelling to the right in the wire

x \leq 0.

K_{2}

is the wave vector of the transmitted wave, then find the ratio of the amplitude of the reflected wave to that of the incident wave.

Solution:
The given incident wave equation is

y = A_{o} s i n (w t - k_{1} x)

where

A_{o}

is the amplitude of the incident wave.
Let

A_{r}

and

A_{t}

are the amplitudes of the reflected and transmitted waves respectively.
Thus the reflected wave is

y_{r} = A_{r} s i n (w t + k_{1} x)

, for

x \leq 0

And the transmitted wave is:

y_{t} = A_{t} s i n (w t - k_{2} x)

, for

x \geq 0

Now applying the boundary conditions: At

x = 0

y + y_{r} = y_{t}

............(1) for all values of

t

⟹ A_{o} + A_{r} = A_{t}

..............(a) (for

t = 0

)
Also, differentiating (1) w.r.t

x

, gives:

\frac{d y}{d x} + \frac{d y_{r}}{d x} = \frac{d y_{t}}{d x}

, for all values of

t

⟹ - k_{1} A_{o} + k_{1} A_{r} = - k_{2} A_{t}

(at

x = 0

and

t = 0

)
Thus

A_{o} - A_{r} = \frac{k_{2}}{k_{1}} A_{t}

.............(b)
From (a) and (b) eliminating

A_{t}

, we get:

\frac{A_{r}}{A_{o}} = \frac{k_{1} - k_{2}}{k_{1} + k_{2}}

Definition

Quincke's Method

This Method's Apparatus is designed for the determination of magnetic susceptibility of a given solution. The apparatus has a pair of electromagnets and a Quincke's Tube in which the sample is taken. This tube has shape U and has two limbs, one with very narrow width compared to the other. The change in the level of the liquid in the narrow limb does not affect the level in the wider limb. The magnetic field is measured using a digital Gauss meter. In this apparatus, the rise in liquid by the application of magnetic field is measured by a travelling microscope. This Method is used to determine magnetic susceptibility of paramagnetic substance in the form of liquid or aqueous solutions.This method is based on the force experienced by a magnetized material in a non-uniform magnetic field. When an object is placed in a magnetic field, a magnetic moment is induced in it. Basically, magnetic susceptibility is a proportionality constant which is dimensionless and indicates the degree of magnetization of a material in response to applied magnetic field. A liquid sample in a narrow tube placed between the poles of a magnet experiences a force and hence when the field is turned on, the meniscus in the narrow tube rises by an amount h, relative to its zero-field position. A measurement of this rise helps to determine the susceptibility of the solution.

Example

Total intensity of two superimposing waves

Example: Two sound waves of equal intensity

I

superimpose at point

P

90^{\circ}

out of phase. What will be the resultant intensity at point

P

?

Solution:
Amplitude of the resultant wave is

A_{R} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos (θ)}

.
Here,

A_{1} = A_{2} = A and θ = π / 2

, so

A_{R} = A \sqrt{2 (1 + \cos (θ))} = 2 A \cos (θ / 2) = 2 A \cos (π / 4) = \sqrt{2} A

I_{R} = | A_{R} |^{2} = 2 A^{2} = 2 I

Example

Write total intensity in terms of component intensities for constructive interference

Two coherent light beams of intensities I and 4I are superposed. Find the maximum and minimum possible intensities in the resulting beam.

Solution:
The net intensity due to interference of two beams of intensities

I_{1}, I_{2}

with a phase difference of

ϕ

I_{n e t} = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} c o s ϕ

Thus,

I_{n e t_{m a x}}

occurs for

c o s ϕ = 1

\Rightarrow I_{n e t_{m a x}} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2}

= 9 I

Similarly,

I_{n e t_{m i n}}

occurs for

c o s ϕ = - 1

\Rightarrow I_{n e t_{m i n}} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2}

= I

Example

Write total intensity in terms of component intensities for destructive interference

Example: Two waves of intensities

I

and

4 I

produce interference. Then find the intensity of constructive and destructive interference.

Solution:

I_{m a x} = (\sqrt{I_{1}} + \sqrt{I_{2}})^{2} = 9 I I_{m i n} = (\sqrt{I_{1}} - \sqrt{I_{2}})^{2} = I

Example

Interference at a given point for sound waves from two given sources

Example: When interference is produced by two progressive waves of equal frequencies, then the maximum intensity of the resulting sound are N times the intensity of each of the component waves. Find the value of N.

Solution: