Question

If $\mu=A+\dfrac{B}{\lambda }+\dfrac{C}{{{\lambda }^{2}}}$ then find the dimensions of $A,B$ and $C$ respectively ($A,B$ and $C$ are constants) where $\lambda$ is wavelength of wave.

Answer 1

We will use the application of dimensional formula to solve the problem. Principle of homogeneity will also be used during the solution. Principle of homogeneity states that dimensions of both sides in a valid dimensional equation must be the same.
Formula used:
We will use the following dimensional formula to solve the problem:-
$\lambda =L$ and $\mu=ML{{T}^{-1}}$

Complete step by step solution:
We have $\mu=A+\dfrac{B}{\lambda }+\dfrac{C}{{{\lambda }^{2}}}$where $A,B$ and $C$ are constants and $\lambda$ is the wavelength of the wave. Considering,$m$ as mass and $u$ as velocity. We get $\mu$ as momentum because momentum is defined as the product of mass and velocity.
We have
$\mu=A+\dfrac{B}{\lambda }+\dfrac{C}{{{\lambda }^{2}}}$………………… $(i)$
We know that dimension of momentum,$\mu=ML{{T}^{-1}}$.
Dimension of wavelength,$\lambda =L$
Putting the dimensions in equation $(i)$ we get
$ML{{T}^{-1}}=A+\dfrac{B}{L}+\dfrac{C}{{{L}^{2}}}$……………….$(ii)$
We will equate $(ii)$ for the constants $A,B$ and $C$ according to the principle of homogeneity respectively.
For $A$ we have
$A=ML{{T}^{-1}}$
Hence, dimensions of $A=ML{{T}^{-1}}$
For $B$ we have
$\dfrac{B}{L}=ML{{T}^{-1}}$
$B=\left[ ML{{T}^{-1}} \right]L$
$B=M{{L}^{2}}{{T}^{-1}}$
Hence, dimensions of $B=M{{L}^{2}}{{T}^{-1}}$.
Now for equating $(ii)$ for $C$ we have
$\dfrac{C}{{{L}^{2}}}=ML{{T}^{-1}}$
$C=\left[ ML{{T}^{-1}} \right]{{L}^{2}}$
$C=M{{L}^{3}}{{T}^{-1}}$
Hence, dimensions of $C=M{{L}^{3}}{{T}^{-1}}$.
Therefore we get the dimensions of the constants $A,B$ and $C$ by basic use of dimensional analysis and principle of homogeneity.

Additional Information:
The study of the relationship between physical quantities by the use of units and dimensions is called dimensional analysis. According to the concept of dimensional analysis if two physical quantities have the same dimensions then they are equal. Dimensional analysis is used to check the consistency of the dimensional equation but this concept does not tell us whether the physical quantity is scalar or vector.

Note:
We should take care of the dimensional formula of every parameter in the problem. We have considered $\mu$ as momentum and solved the problem. We can also consider $\mu$ as constant and solve the problem accordingly.