Dispersion Relations in Plasma MCQ Quiz - Objective Question with Answer for Dispersion Relations in Plasma - Download Free PDF
Last updated on Apr 21, 2025
Latest Dispersion Relations in Plasma MCQ Objective Questions
Dispersion Relations in Plasma Question 1:
Taking the electronic charge as \(e\) and the permittivity as \(\epsilon_{0}\), use dimensional analysis to determine the correct expression for \(\omega _{p}\).
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 1 Detailed Solution
Calculation:
We have ωp as the angular frequency.
Thus its unit would be given as [T-1]
We have the units of density (N) as 1/L3, charge (e) as IT, mass as M, and ε0 is evaluated using the force equation F = 1 / (4πε0) × (q1q2 / r2) as (I2T4) / (ML3).
By evaluating the units of each given option we get:
From dimension analysis: [T]-1 = [N]x[e]y[M]z[ε]t ...(i)
Substituting the units: [T]-1 = [1/L3]x[IT]y[M]z[(I2T4) / (ML3)]t.
Comparing the powers of M, L, I, and T:
z - t = 0 ...(ii)
-3x - 3t = 0 ...(iii)
y + 2t = 0 ...(iv)
y + 4t = -1 ...(v)
Solving equations: y = 1; t = -1/2; z = -1/2; x = 1/2.
Substituting x, y, z, and t in equation (i): ωp = &sqrt;(N) e &sqrt;(m) &sqrt;(ε) = &sqrt;(Ne2 / mε).
Dispersion Relations in Plasma Question 2:
Taking the electronic charge as \(e\) and the permittivity as \(\epsilon_{0}\), use dimensional analysis to determine the correct expression for \(\omega _{p}\).
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 2 Detailed Solution
Calculation:
We have ωp as the angular frequency.
Thus its unit would be given as [T-1]
We have the units of density (N) as 1/L3, charge (e) as IT, mass as M, and ε0 is evaluated using the force equation F = 1 / (4πε0) × (q1q2 / r2) as (I2T4) / (ML3).
By evaluating the units of each given option we get:
From dimension analysis: [T]-1 = [N]x[e]y[M]z[ε]t ...(i)
Substituting the units: [T]-1 = [1/L3]x[IT]y[M]z[(I2T4) / (ML3)]t.
Comparing the powers of M, L, I, and T:
z - t = 0 ...(ii)
-3x - 3t = 0 ...(iii)
y + 2t = 0 ...(iv)
y + 4t = -1 ...(v)
Solving equations: y = 1; t = -1/2; z = -1/2; x = 1/2.
Substituting x, y, z, and t in equation (i): ωp = &sqrt;(N) e &sqrt;(m) &sqrt;(ε) = &sqrt;(Ne2 / mε).
Dispersion Relations in Plasma Question 3:
Suppose that there is a dispersive medium whose refractive index depends on the wavelength as given by \(n(\lambda)=n_0+\frac{a}{\lambda^2}-\frac{b}{\lambda^4}\). The value of 𝜆 at which the group and phase velocities would be the same, is:
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 3 Detailed Solution
Concept :
The problem involves the relationship between group velocity and phase velocity in a dispersive medium, where the refractive index n(\(\lambda\)) depends on the wavelength \(\lambda \) . The group velocity \(v_g\) is the velocity at which the envelope of the wave packet travels, while the phase velocity \(v_p\) is the velocity at which individual wave crests move. The condition where the group and phase velocities are equal provides a critical insight into the dispersion relation.
The refractive index is given by:
\(\ n(\lambda) = n_0 + \frac{a}{\lambda^2} - \frac{b}{\lambda^4}\)
Solution :
To find the value of \lambda where the group velocity and phase velocity are equal, we use the relationship between the two. The phase velocity \(v_p\) is given by:
\(\ v_p = \frac{c}{n(\lambda)}\)
where c is the speed of light. The group velocity \(v_g\) is given by:
\(\ v_g = v_p - \lambda \frac{dv_p}{d\lambda} \)
We need to set \(v_g = v_p\) , and \(v_p=\frac{c}{n} \\ \) so:
\(- \lambda \frac{dv_p}{d\lambda} = 0\)
Taking the derivative of the refractive index with respect to \(\lambda :\)
\(\ \frac{dn}{d\lambda} = -\frac{2a}{\lambda^3} + \frac{4b}{\lambda^5}\)
Now, substitute this \( \frac{dv_p}{d\lambda}=-\frac{c}{n^2}\frac{dn}{d\lambda}\)
into the equation,\(- \lambda \frac{dv_p}{d\lambda} = 0\)
\(\ \frac{dn}{d\lambda} = -\frac{2a}{\lambda^3} + \frac{4b}{\lambda^5} =0\)
\(\lambda = \sqrt{\frac{2b}{a}}\)
The correct option is (1) : \(\lambda =\sqrt{\frac{2b}{a}}\)
Top Dispersion Relations in Plasma MCQ Objective Questions
Dispersion Relations in Plasma Question 4:
Taking the electronic charge as \(e\) and the permittivity as \(\epsilon_{0}\), use dimensional analysis to determine the correct expression for \(\omega _{p}\).
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 4 Detailed Solution
Calculation:
We have ωp as the angular frequency.
Thus its unit would be given as [T-1]
We have the units of density (N) as 1/L3, charge (e) as IT, mass as M, and ε0 is evaluated using the force equation F = 1 / (4πε0) × (q1q2 / r2) as (I2T4) / (ML3).
By evaluating the units of each given option we get:
From dimension analysis: [T]-1 = [N]x[e]y[M]z[ε]t ...(i)
Substituting the units: [T]-1 = [1/L3]x[IT]y[M]z[(I2T4) / (ML3)]t.
Comparing the powers of M, L, I, and T:
z - t = 0 ...(ii)
-3x - 3t = 0 ...(iii)
y + 2t = 0 ...(iv)
y + 4t = -1 ...(v)
Solving equations: y = 1; t = -1/2; z = -1/2; x = 1/2.
Substituting x, y, z, and t in equation (i): ωp = &sqrt;(N) e &sqrt;(m) &sqrt;(ε) = &sqrt;(Ne2 / mε).
Dispersion Relations in Plasma Question 5:
Taking the electronic charge as \(e\) and the permittivity as \(\epsilon_{0}\), use dimensional analysis to determine the correct expression for \(\omega _{p}\).
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 5 Detailed Solution
Calculation:
We have ωp as the angular frequency.
Thus its unit would be given as [T-1]
We have the units of density (N) as 1/L3, charge (e) as IT, mass as M, and ε0 is evaluated using the force equation F = 1 / (4πε0) × (q1q2 / r2) as (I2T4) / (ML3).
By evaluating the units of each given option we get:
From dimension analysis: [T]-1 = [N]x[e]y[M]z[ε]t ...(i)
Substituting the units: [T]-1 = [1/L3]x[IT]y[M]z[(I2T4) / (ML3)]t.
Comparing the powers of M, L, I, and T:
z - t = 0 ...(ii)
-3x - 3t = 0 ...(iii)
y + 2t = 0 ...(iv)
y + 4t = -1 ...(v)
Solving equations: y = 1; t = -1/2; z = -1/2; x = 1/2.
Substituting x, y, z, and t in equation (i): ωp = &sqrt;(N) e &sqrt;(m) &sqrt;(ε) = &sqrt;(Ne2 / mε).
Dispersion Relations in Plasma Question 6:
Taking the electronic charge as \(e\) and the permittivity as \(\epsilon_{0}\), use dimensional analysis to determine the correct expression for \(\omega _{p}\).
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 6 Detailed Solution
Calculation:
We have ωp as the angular frequency.
Thus its unit would be given as [T-1]
We have the units of density (N) as 1/L3, charge (e) as IT, mass as M, and ε0 is evaluated using the force equation F = 1 / (4πε0) × (q1q2 / r2) as (I2T4) / (ML3).
By evaluating the units of each given option we get:
From dimension analysis: [T]-1 = [N]x[e]y[M]z[ε]t ...(i)
Substituting the units: [T]-1 = [1/L3]x[IT]y[M]z[(I2T4) / (ML3)]t.
Comparing the powers of M, L, I, and T:
z - t = 0 ...(ii)
-3x - 3t = 0 ...(iii)
y + 2t = 0 ...(iv)
y + 4t = -1 ...(v)
Solving equations: y = 1; t = -1/2; z = -1/2; x = 1/2.
Substituting x, y, z, and t in equation (i): ωp = &sqrt;(N) e &sqrt;(m) &sqrt;(ε) = &sqrt;(Ne2 / mε).
Dispersion Relations in Plasma Question 7:
Suppose that there is a dispersive medium whose refractive index depends on the wavelength as given by \(n(\lambda)=n_0+\frac{a}{\lambda^2}-\frac{b}{\lambda^4}\). The value of 𝜆 at which the group and phase velocities would be the same, is:
Answer (Detailed Solution Below)
Dispersion Relations in Plasma Question 7 Detailed Solution
Concept :
The problem involves the relationship between group velocity and phase velocity in a dispersive medium, where the refractive index n(\(\lambda\)) depends on the wavelength \(\lambda \) . The group velocity \(v_g\) is the velocity at which the envelope of the wave packet travels, while the phase velocity \(v_p\) is the velocity at which individual wave crests move. The condition where the group and phase velocities are equal provides a critical insight into the dispersion relation.
The refractive index is given by:
\(\ n(\lambda) = n_0 + \frac{a}{\lambda^2} - \frac{b}{\lambda^4}\)
Solution :
To find the value of \lambda where the group velocity and phase velocity are equal, we use the relationship between the two. The phase velocity \(v_p\) is given by:
\(\ v_p = \frac{c}{n(\lambda)}\)
where c is the speed of light. The group velocity \(v_g\) is given by:
\(\ v_g = v_p - \lambda \frac{dv_p}{d\lambda} \)
We need to set \(v_g = v_p\) , and \(v_p=\frac{c}{n} \\ \) so:
\(- \lambda \frac{dv_p}{d\lambda} = 0\)
Taking the derivative of the refractive index with respect to \(\lambda :\)
\(\ \frac{dn}{d\lambda} = -\frac{2a}{\lambda^3} + \frac{4b}{\lambda^5}\)
Now, substitute this \( \frac{dv_p}{d\lambda}=-\frac{c}{n^2}\frac{dn}{d\lambda}\)
into the equation,\(- \lambda \frac{dv_p}{d\lambda} = 0\)
\(\ \frac{dn}{d\lambda} = -\frac{2a}{\lambda^3} + \frac{4b}{\lambda^5} =0\)
\(\lambda = \sqrt{\frac{2b}{a}}\)
The correct option is (1) : \(\lambda =\sqrt{\frac{2b}{a}}\)