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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 ρ 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 S1/n. The value of n is:

Solution:

Problem is based on equality of dimension.
Step 1: As we don't know any relation between known [ρ,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 = ρx × Sy × fz .. (1)
Step 2: Express the data in its fundamental (base) quantity
Distance [d] = [L]
Density [ρ] = [ML-3]
Intensity [S] = [MT-3]
Frequency [f] = [T-1]
Step 3: By equality of dimension analysis, dimensions on LHS should be equal to that in RHS
[L] = [ML-3]x[MT-3]y[T-1]z
[L] = [Mx+y][L-3x][T-3y-z]
aranging
On equating LHS and RHS, we get,
x + y = 0
-3x = 1
-3y - z = 0
x = -1/3 y = 1/3 z = -1
Substitution in (1) yields,
d = ρ-1/3 × S1/3 × f-1
Step 4: Comparison with S1/n with S1/3 gives:
n = 3