Question

To find the distance $d$ over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog, intensity (power / area) $S$ of the light from the signal and its frequency $f$. The engineer finds that $d$ is proportional to $S$${}^{1/n}$. The value of $n$ is :

Answer 1

Problem is based on equality of dimension.
Step 1: As we don't know any relation between known [$\rho ,S,f$] and unknown [$d$] quantity, lets us take unknown on LHS and equate it with known on RHS, with each factor in product form and raised to some arbitrary powers say $x,y$ & $z$
$d={\rho }^x\times S^y\times f^z$ .......(1)
Step 2: Express the data in its fundamental (base) quantity
Distance $\left[d\right]=\left[L\right]$
Density $\left[\rho \right]=\left[M{L}^{-3}\right]$
Intensity $\left[S\right]=\left[M{T}^{-3}\right]$
Frequency$\left[f\right]=\left[T^{-1}\right]$
Step 3: By equality of dimension analysis, dimensions on LHS should be equal to that in RHS
$\left[L\right]=[M{L}^{-3}{]}^x{\left[M{T}^{-3}\right]}^y{\left[T^{-1}\right]}^z$
$\left[L\right]={[M}^{x+y}\left]\right[L^{-3x}\left]\right[T^{-3y-z}]$ ............arranging
On equating LHS and RHS, we get,
$x+y=0-3x=1\&-3y-z=0$
$x=\frac{-1}{3}y=\frac{1}{3}\&z=-1$
Substitution in (1) yields,
$d={\rho }^{-1/3}\times S^{1/3}\times f^{-1}$
Step 4: Comparison with $S^{1/n}$ with $S^{1/3}$ gives:
$n=3$