Q. The refractive index of a material is given by the equation $n = \frac{A\ + B}{\lambda^{2}},$ where $A\ and\ B$ are constant. The dimensional formula for $B$ is
A) $\left\lbrack M^{0}L^{2}T^{- 1} \right\rbrack$
B) $\left\lbrack M^{0}L^{- 2}T^{0} \right\rbrack$
C) $\left\lbrack M^{0}L^{2}T^{- 2} \right\rbrack$
D) $\left\lbrack M^{0}L^{2}T^{0} \right\rbrack$
Explanation
The given equation is $\ \ \ \ \ \ \ \ \ \ \ \ \ \ n = \frac{A + B}{\lambda^{2}}$ Where $A\ and\ B$ are constants. By homogeneity principle the dimensions of all the terms on both sides should be same $ie,\ \ \ \ \ \lbrack B\rbrack = \lbrack A\rbrack = \left\lbrack n\lambda^{2} \right\rbrack$ $\therefore\ \ \ \ \ \ \lbrack B\rbrack = \left\lbrack M^{0}L^{0}T^{0} \right\rbrack\left\lbrack L^{2} \right\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left\lbrack M^{0}L^{2}T^{0} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.