Assume the dependence

Let the visible distance $d$ depend on mass density of fog $\rho$, light intensity $S$, and frequency $f$ as $d \propto \rho^a S^b f^c$

We need to find the exponent of $S$, i.e. $b$, and then compare with $d \propto S^{1/n}$

Write dimensions of each quantity

• Distance: $[d] = L$

• Mass density: $[\rho] = M L^{-3}$

• Intensity (power/area):

  Power $= \dfrac{\text{energy}}{\text{time}}$, and $[\text{energy}] = M L^2 T^{-2}$ so $[\text{power}] = M L^2 T^{-3}$ Hence intensity $[S] = \frac{M L^2 T^{-3}}{L^2} = M T^{-3}$

• Frequency: $[f] = T^{-1}$

Substitute dimensions

$[d] = [\rho]^a [S]^b [f]^c$ $L = (M L^{-3})^a (M T^{-3})^b (T^{-1})^c$

Simplifying, $L = M^{a+b} L^{-3a} T^{-3b-c}$

Equate powers of fundamental dimensions

Comparing both sides:

• For $M$: $a+b=0$
• For $L$: $-3a=1$
• For $T$: $-3b-c=0$

Solve for $a,b,c$

From $-3a=1 \Rightarrow a=-\frac{1}{3}$

Then from $a+b=0 \Rightarrow b=\frac{1}{3}$

So, $d \propto S^{1/3}$

Compare with the given form

Since $d \propto S^{1/n}$ and we found $d \propto S^{1/3}$ therefore, $n=3$

Comparison with stored answer

Stored correct answer = $3$

Our derived answer also gives $3$, so they agree.