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IGNOU BPHCT-137 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BPHCT-137 Assignment Question Paper 2024

bphct-137-solved-assignment-2024-qp-ff2bf642-4456-4ac1-9a8c-9a60ad6420c6

bphct-137-solved-assignment-2024-qp-ff2bf642-4456-4ac1-9a8c-9a60ad6420c6

BPHCT-137 Solved Assignment 2024 QP
PART A
  1. a) Obtain an expression for energy transported by a progressive wave on a taut string. Also show that the power transported by the wave is proportional to the wave velocity.
    b) The fundamental frequency of a string instrument is 580 H z 580 H z 580Hz580 \mathrm{~Hz}580 Hz. (i) Calculate the frequency of the first and third harmonics generated on it. (ii) If the tension in the string is doubled, calculate the new fundamental frequency.
    c) Show that when two in-phase linearly polarised light waves are superposed, the resultant wave has fixed orientation as well as amplitude. Depict the orientation of electric field vector of the resultant wave in the reference plane.
  2. a) Explain how Michelson interferometer can be used to determine the wavelength of light?
    b) In a Young’s double slit experiment, a monochromatic light of wavelength 600 n m 600 n m 600nm600 \mathrm{~nm}600 nm is used. The slits are 0.2 m m 0.2 m m 0.2mm0.2 \mathrm{~mm}0.2 mm apart. An interference pattern is observed on a screen 0.6 m 0.6 m 0.6m0.6 \mathrm{~m}0.6 m away. Calculate the distance between the central maxima and the third minima on the screen.
    c) A Michelson interferometer is illuminated with monochromatic light. When one of the mirrors is moved 2.5 × 10 − 5 m , 100 2.5 × 10 − 5 m , 100 2.5 xx10^(-5)m,1002.5 \times 10^{-5} \mathrm{~m}, 1002.5×10−5 m,100 fringes cross the field of view. Determine the wavelength of the incident light.
    d) Obtain an expression of the displacement of the nth bright fringe in double slit experiment when a thin transparent plate of thickness t t ttt and refractive index μ μ mu\muμ is introduced in the path of one of the two interfering waves. How the fringe-width is affected with the introduction of the plate?
    e) Interference fringes are formed by reflection from a thin air wedge by using sodium light of wavelength of 5893 Å 5893 Å 5893″Å”5893 \AA5893Å When viewed perpendicularly, 12 fringes are observed in a distance of 1 c m 1 c m 1cm1 \mathrm{~cm}1 cm. Calculate the angle of wedge.
PART B
3. a) In the experimental set up to observe diffraction pattern of a straight edge, the distance of the edge from the source is 30 c m 30 c m 30cm30 \mathrm{~cm}30 cm and the distance of the screen from the edge is 40 c m 40 c m 40cm40 \mathrm{~cm}40 cm. The wavelength of the light used is 480 × 10 − 5 c m 480 × 10 − 5 c m 480 xx10^(-5)cm480 \times 10^{-5} \mathrm{~cm}480×10−5 cm. Calculate the position of the first and third minima from the edge of the geometrical shadow.
b) Obtain an expression for intensity distribution in a single slit diffraction pattern.
c) A plane light wave of wavelength 580 n m 580 n m 580nm580 \mathrm{~nm}580 nm falls on a long narrow slit of width 0.5 m m 0.5 m m 0.5mm0.5 \mathrm{~mm}0.5 mm. (i) Calculate the angles of diffraction for the first two minima. (ii) How are these angles influenced if the width of slit is changed to 0.2 m m 0.2 m m 0.2mm0.2 \mathrm{~mm}0.2 mm ? (iii) If a convex lens of focal length 0.15 m 0.15 m 0.15m0.15 \mathrm{~m}0.15 m is now placed after the slit, calculate the separation between the second minima on either side of the central maximum.
4. a) Explain the three light-matter interactions with the help of atomic level diagrams. Write the corresponding Einstein’s equations.
b) Explain the role of resonant cavity in determining the coherence and monochromaticity of a laser light.
c) Describe the types of optical fibres. How can the pulse dispersion in the optical fibre be reduces?
d) A laser cavity of 25 c m 25 c m 25cm25 \mathrm{~cm}25 cm length sustains the radiation of 6300 A 6300 A 6300A6300 \mathrm{~A}6300 A. Calculate the number of modes and mode separation. If the spread in wavelength is 0.02 Å 0.02 Å 0.02″Å”0.02 \AA0.02Å, calculate the coherence length and coherence time.
e) The refractive indices of core ( n 1 ) n 1 (n_(1))\left(n_1\right)(n1) and cladding ( n 2 ) n 2 (n_(2))\left(n_2\right)(n2) materials of two optical fibres A A AAA and B B BBB are as follows:
( n 1 ) A = 1.46 and ( n 2 ) A = 1.38 n 1 A = 1.46 and n 2 A = 1.38 (n_(1))_(A)=1.46″ and “(n_(2))_(A)=1.38\left(n_1\right)_A=1.46 \text { and }\left(n_2\right)_A=1.38(n1)A=1.46 and (n2)A=1.38
and ( n 1 ) B = 1.52 n 1 B = 1.52 quad(n_(1))_(B)=1.52\quad\left(n_1\right)_B=1.52(n1)B=1.52 and ( n 2 ) B = 1.41 n 2 B = 1.41 (n_(2))_(B)=1.41\left(n_2\right)_B=1.41(n2)B=1.41
Which of the two fibres will have higher gathering capacity?
\(b^2=c^2+a^2-2ac\:Cos\left(B\right)\)

BPHCT-137 Sample Solution 2024

bphct-137-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

bphct-137-solved-assignment-2024-ss-b0a87631-f2d0-400e-939b-0b99078c2d98

BPHCT-137 Solved Assignment 2024 SS
PART A
  1. a) Obtain an expression for energy transported by a progressive wave on a taut string. Also show that the power transported by the wave is proportional to the wave velocity.
Answer:
The energy transported by a traveling wave on a taut string can be analyzed by considering the kinetic and potential energies of a small element of the string. Let’s derive the expressions for these energies and then calculate the total energy and power transported by the wave.

Kinetic Energy:

For a small element Δ x Δ x Delta x\Delta xΔx of the string, the kinetic energy ( d ( K . E . ) d ( K . E . ) d(K.E.)d(K.E.)d(K.E.)) is given by:
d ( K . E . ) = 1 2 m v 2 = 1 2 ( ρ Δ x ) ( ∂ y ∂ t ) 2 d ( K . E . ) = 1 2 m v 2 = 1 2 ( ρ Δ x ) ∂ y ∂ t 2 d(K.E.)=(1)/(2)mv^(2)=(1)/(2)(rho Delta x)((del y)/(del t))^(2)d(K.E.) = \frac{1}{2} m v^2 = \frac{1}{2} (\rho \Delta x) \left(\frac{\partial y}{\partial t}\right)^2d(K.E.)=12mv2=12(ρΔx)(∂y∂t)2
where ρ ρ rho\rhoρ is the mass per unit length of the string, and ∂ y ∂ t ∂ y ∂ t (del y)/(del t)\frac{\partial y}{\partial t}∂y∂t is the velocity of the element.
For a sinusoidal wave y ( x , t ) = a sin ⁡ ( ω t − k x ) y ( x , t ) = a sin ⁡ ( ω t − k x ) y(x,t)=a sin(omega t-kx)y(x, t) = a \sin(\omega t – kx)y(x,t)=asin⁡(ωt−kx), the velocity is:
∂ y ∂ t = a ω cos ⁡ ( ω t − k x ) ∂ y ∂ t = a ω cos ⁡ ( ω t − k x ) (del y)/(del t)=a omega cos(omega t-kx)\frac{\partial y}{\partial t} = a \omega \cos(\omega t – kx)∂y∂t=aωcos⁡(ωt−kx)
Substituting this into the kinetic energy expression:
d ( K . E . ) = 1 2 ( ρ Δ x ) [ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) ] d ( K . E . ) = 1 2 ( ρ Δ x ) a 2 ω 2 cos 2 ⁡ ( ω t − k x ) d(K.E.)=(1)/(2)(rho Delta x)[a^(2)omega^(2)cos^(2)(omega t-kx)]d(K.E.) = \frac{1}{2} (\rho \Delta x) \left[a^2 \omega^2 \cos^2(\omega t – kx)\right]d(K.E.)=12(ρΔx)[a2ω2cos2⁡(ωt−kx)]

Potential Energy:

The potential energy ( d ( P . E . ) d ( P . E . ) d(P.E.)d(P.E.)d(P.E.)) arises due to the stretching of the string and is given by:
d ( P . E . ) = 1 2 T ( ∂ y ∂ x ) 2 Δ x d ( P . E . ) = 1 2 T ∂ y ∂ x 2 Δ x d(P.E.)=(1)/(2)T((del y)/(del x))^(2)Delta xd(P.E.) = \frac{1}{2} T \left(\frac{\partial y}{\partial x}\right)^2 \Delta xd(P.E.)=12T(∂y∂x)2Δx
where T T TTT is the tension in the string.
For the sinusoidal wave, the derivative with respect to x x xxx is:
∂ y ∂ x = − a k cos ⁡ ( ω t − k x ) ∂ y ∂ x = − a k cos ⁡ ( ω t − k x ) (del y)/(del x)=-ak cos(omega t-kx)\frac{\partial y}{\partial x} = -a k \cos(\omega t – kx)∂y∂x=−akcos⁡(ωt−kx)
Substituting this into the potential energy expression:
d ( P . E . ) = 1 2 ( T Δ x ) [ a 2 k 2 cos 2 ⁡ ( ω t − k x ) ] d ( P . E . ) = 1 2 ( T Δ x ) a 2 k 2 cos 2 ⁡ ( ω t − k x ) d(P.E.)=(1)/(2)(T Delta x)[a^(2)k^(2)cos^(2)(omega t-kx)]d(P.E.) = \frac{1}{2} (T \Delta x) \left[a^2 k^2 \cos^2(\omega t – kx)\right]d(P.E.)=12(TΔx)[a2k2cos2⁡(ωt−kx)]

Total Energy:

The total mechanical energy ( d E d E dEdEdE) of the element Δ x Δ x Delta x\Delta xΔx is the sum of the kinetic and potential energies:
d E = d ( K . E . ) + d ( P . E . ) = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) [ ρ ω 2 + T k 2 ] = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) [ 2 ρ ω 2 ] (since v = ω k and T = ρ v 2 ) = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) Δ x d E = d ( K . E . ) + d ( P . E . ) = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) ρ ω 2 + T k 2 = 1 2 Δ x a 2 cos 2 ⁡ ( ω t − k x ) 2 ρ ω 2 (since v = ω k and T = ρ v 2 ) = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) Δ x {:[dE=d(K.E.)+d(P.E.)],[=(1)/(2)Delta xa^(2)cos^(2)(omega t-kx)[rhoomega^(2)+Tk^(2)]],[=(1)/(2)Delta xa^(2)cos^(2)(omega t-kx)[2rhoomega^(2)]quad(since v=(omega )/(k)” and “T=rhov^(2))],[=rhoa^(2)omega^(2)cos^(2)(omega t-kx)Delta x]:}\begin{aligned} dE &= d(K.E.) + d(P.E.) \\ &= \frac{1}{2} \Delta x a^2 \cos^2(\omega t – kx) \left[\rho \omega^2 + T k^2\right] \\ &= \frac{1}{2} \Delta x a^2 \cos^2(\omega t – kx) \left[2 \rho \omega^2\right] \quad \text{(since } v = \frac{\omega}{k} \text{ and } T = \rho v^2) \\ &= \rho a^2 \omega^2 \cos^2(\omega t – kx) \Delta x \end{aligned}dE=d(K.E.)+d(P.E.)=12Δxa2cos2⁡(ωt−kx)[ρω2+Tk2]=12Δxa2cos2⁡(ωt−kx)[2ρω2](since v=ωk and T=ρv2)=ρa2ω2cos2⁡(ωt−kx)Δx
The linear energy density ( D ( x , t ) D ( x , t ) D(x,t)D(x, t)D(x,t)) of the wave is then:
D ( x , t ) = d E Δ x = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) D ( x , t ) = d E Δ x = ρ a 2 ω 2 cos 2 ⁡ ( ω t − k x ) D(x,t)=(dE)/(Delta x)=rhoa^(2)omega^(2)cos^(2)(omega t-kx)D(x, t) = \frac{dE}{\Delta x} = \rho a^2 \omega^2 \cos^2(\omega t – kx)D(x,t)=dEΔx=ρa2ω2cos2⁡(ωt−kx)

Energy Contained in One Wavelength:

The total mechanical energy ( E E EEE) contained in one wavelength ( λ λ lambda\lambdaλ) of the wave is obtained by integrating D ( x , t ) D ( x , t ) D(x,t)D(x, t)D(x,t) over one wavelength:
E = ∫ 0 λ D ( x , t ) d x = ρ a 2 ω 2 ∫ 0 λ cos 2 ⁡ ( k x ) d x = 1 2 ρ a 2 ω 2 λ E = ∫ 0 λ   D ( x , t ) d x = ρ a 2 ω 2 ∫ 0 λ   cos 2 ⁡ ( k x ) d x = 1 2 ρ a 2 ω 2 λ E=int_(0)^(lambda)D(x,t)dx=rhoa^(2)omega^(2)int_(0)^(lambda)cos^(2)(kx)dx=(1)/(2)rhoa^(2)omega^(2)lambdaE = \int_0^\lambda D(x, t) dx = \rho a^2 \omega^2 \int_0^\lambda \cos^2(kx) dx = \frac{1}{2} \rho a^2 \omega^2 \lambdaE=∫0λD(x,t)dx=ρa2ω2∫0λcos2⁡(kx)dx=12ρa2ω2λ

Power Transported by the Wave:

The power ( P P PPP) of the wave is the energy averaged over one time period ( T T TTT) and is given by:
P = E T = E ( λ / v ) = 1 2 ρ a 2 ω 2 v = 2 π 2 ρ a 2 f 2 v P = E T = E ( λ / v ) = 1 2 ρ a 2 ω 2 v = 2 π 2 ρ a 2 f 2 v P=(E)/(T)=(E)/((lambda//v))=(1)/(2)rhoa^(2)omega^(2)v=2pi^(2)rhoa^(2)f^(2)vP = \frac{E}{T} = \frac{E}{(\lambda / v)} = \frac{1}{2} \rho a^2 \omega^2 v = 2 \pi^2 \rho a^2 f^2 vP=ET=E(λ/v)=12ρa2ω2v=2π2ρa2f2v
where ω = 2 π f ω = 2 π f omega=2pi f\omega = 2 \pi fω=2πf and f f fff is the frequency of the wave.

Conclusion:

The energy transported by a traveling wave on a taut string is distributed between kinetic and potential energies, with the total energy contained in one wavelength given by 1 2 ρ a 2 ω 2 λ 1 2 ρ a 2 ω 2 λ (1)/(2)rhoa^(2)omega^(2)lambda\frac{1}{2} \rho a^2 \omega^2 \lambda12ρa2ω2λ. The power transported by the wave is proportional to the wave velocity, as well as the square of its amplitude and frequency.
b) The fundamental frequency of a string instrument is 580 H z 580 H z 580Hz580 \mathrm{~Hz}580 Hz. (i) Calculate the frequency of the first and third harmonics generated on it. (ii) If the tension in the string is doubled, calculate the new fundamental frequency.
Answer:
To address this problem, we’ll break it down into two parts as requested:

Part (i): Frequency of the First and Third Harmonics

The fundamental frequency of a string instrument is given as f 1 = 580 H z f 1 = 580 H z f_(1)=580Hzf_1 = 580 \, \mathrm{Hz}f1=580Hz. The harmonics of a string instrument are integer multiples of the fundamental frequency. Thus, the frequency of the n n nnnth harmonic can be calculated using the formula:
f n = n ⋅ f 1 f n = n ⋅ f 1 f_(n)=n*f_(1)f_n = n \cdot f_1fn=n⋅f1
  • For the first harmonic ( n = 1 n = 1 n=1n=1n=1), this is simply the fundamental frequency itself, f 1 = 580 H z f 1 = 580 H z f_(1)=580Hzf_1 = 580 \, \mathrm{Hz}f1=580Hz.
  • For the third harmonic ( n = 3 n = 3 n=3n=3n=3), the frequency is f 3 = 3 ⋅ f 1 f 3 = 3 ⋅ f 1 f_(3)=3*f_(1)f_3 = 3 \cdot f_1f3=3⋅f1.
Let’s substitute the values to find f 3 f 3 f_(3)f_3f3:
f 3 = 3 ⋅ 580 H z f 3 = 3 ⋅ 580 H z f_(3)=3*580Hzf_3 = 3 \cdot 580 \, \mathrm{Hz}f3=3⋅580Hz
After Calculating, we get:
f 3 = 1740 H z f 3 = 1740 H z f_(3)=1740Hzf_3 = 1740 \, \mathrm{Hz}f3=1740Hz

Part (ii): New Fundamental Frequency if the Tension is Doubled

The fundamental frequency of a string can also be expressed in terms of the tension ( T T TTT) in the string, its length ( L L LLL), and its mass per unit length ( μ μ mu\muμ) as follows:
f 1 = 1 2 L T μ f 1 = 1 2 L T μ f_(1)=(1)/(2L)sqrt((T)/( mu))f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}f1=12LTμ
When the tension in the string is doubled ( T ′ = 2 T T ′ = 2 T T^(‘)=2TT’ = 2TT′=2T), the new fundamental frequency ( f 1 ′ f 1 ′ f_(1)^(‘)f_1′f1′) can be calculated using the modified tension:
f 1 ′ = 1 2 L T ′ μ = 1 2 L 2 T μ f 1 ′ = 1 2 L T ′ μ = 1 2 L 2 T μ f_(1)^(‘)=(1)/(2L)sqrt((T^(‘))/(mu))=(1)/(2L)sqrt((2T)/(mu))f_1′ = \frac{1}{2L} \sqrt{\frac{T’}{\mu}} = \frac{1}{2L} \sqrt{\frac{2T}{\mu}}f1′=12LT′μ=12L2Tμ
Since 2 T = 2 ⋅ T 2 T = 2 ⋅ T sqrt(2T)=sqrt2*sqrtT\sqrt{2T} = \sqrt{2} \cdot \sqrt{T}2T=2⋅T, we can relate the new fundamental frequency to the old one as follows:
f 1 ′ = 2 ⋅ f 1 f 1 ′ = 2 ⋅ f 1 f_(1)^(‘)=sqrt2*f_(1)f_1′ = \sqrt{2} \cdot f_1f1′=2⋅f1
Substituting the given fundamental frequency ( f 1 = 580 H z f 1 = 580 H z f_(1)=580Hzf_1 = 580 \, \mathrm{Hz}f1=580Hz) into the formula gives us:
f 1 ′ = 2 ⋅ 580 H z f 1 ′ = 2 ⋅ 580 H z f_(1)^(‘)=sqrt2*580Hzf_1′ = \sqrt{2} \cdot 580 \, \mathrm{Hz}f1′=2⋅580Hz
After Calculating, we get:
f 1 ′ ≈ 819.6 H z f 1 ′ ≈ 819.6 H z f_(1)^(‘)~~819.6Hzf_1′ \approx 819.6 \, \mathrm{Hz}f1′≈819.6Hz

Summary

  • The frequency of the first harmonic is 580 H z 580 H z 580Hz580 \, \mathrm{Hz}580Hz, which is the same as the fundamental frequency.
  • The frequency of the third harmonic is 1740 H z 1740 H z 1740Hz1740 \, \mathrm{Hz}1740Hz.
  • If the tension in the string is doubled, the new fundamental frequency becomes approximately 819.6 H z 819.6 H z 819.6Hz819.6 \, \mathrm{Hz}819.6Hz.
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