To find the distance $d$ over which a signal can be seen clearly in foggy conditions, a railways-engineer uses dimensions 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

Step 1: Set up the dimensional relationship.

Assume the distance $d$ depends on the mass density $\rho$, intensity $S$, and frequency $f$ via a power-law relation.

$$d = k \rho^{a} S^{b} f^{c}$$

where $k$ is a dimensionless constant.

Step 2: Determine the dimensions of each physical quantity.

Write the dimensional formulas for distance, mass density, intensity, and frequency.

$$[d] = \text{L}$$ $$[\rho] = \text{M}\text{L}^{-3}$$ $$[S] = \frac{\text{Power}}{\text{Area}} = \frac{\text{M}\text{L}^{2}\text{T}^{-3}}{\text{L}^{2}} = \text{M}\text{T}^{-3}$$ $$[f] = \text{T}^{-1}$$

Step 3: Substitute the dimensions into the relation.

Equate the dimensions of both sides of the equation.

$$\text{M}^{0}\text{L}^{1}\text{T}^{0} = (\text{M}\text{L}^{-3})^{a} (\text{M}\text{T}^{-3})^{b} (\text{T}^{-1})^{c}$$ $$\text{M}^{0}\text{L}^{1}\text{T}^{0} = \text{M}^{a+b} \text{L}^{-3a} \text{T}^{-3b-c}$$

Step 4: Set up the system of equations for the exponents.

Equate the powers of $\text{M}$, $\text{L}$, and $\text{T}$ on both sides.

$$a + b = 0$$ $$-3a = 1$$ $$-3b - c = 0$$

Step 5: Solve for the exponents.

Solve the system of equations starting with the equation for $\text{L}$.

$$a = -\frac{1}{3}$$ $$b = -a = \frac{1}{3}$$ $$c = -3b = -1$$

Step 6: Find the value of $n$.

Substitute the exponent $b$ back into the proportionality relation for $d$ and $S$.

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

Comparing this to the given relation $d \propto S^{1/n}$ yields the value of $n$.

$$\frac{1}{n} = \frac{1}{3}$$ $$n = 3$$

Answer: $n = 3$