Physics

If $$\mu =A+\dfrac { B }{ \lambda  } +\dfrac { C }{ { \lambda }^{ 2 } } $$ is dimensionally correct then the dimensions of A, B and C respectively are:
($$\mu $$, A, B, C are constants) 


ANSWER

No dimensions, $$ L, { L }^{ 2 }$$


SOLUTION

$$ \underset{M^{0}L^{0}T^{0}}{\mu}=A+\dfrac{B}{\lambda }+\dfrac{C}{\lambda^{2}}$$

$$ A = No \ dimension = M^{0}L^{0}T^{0} $$

$$\mu=\dfrac{B}{\lambda}\Rightarrow B=\mu \lambda =M^{0}L^{1}T^{0} $$
Similarly, $$C=M^{0}L^{2}T^{0}$$
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