To find the distance d over which a signal can be seen clearly in foggy conditions, a railway engine

Q: Why do we assume the relation is a product of powers?

In physics, most fundamental relationships can be approximated as power laws (like $F=ma$, $K=1/2 mv^2$). Dimensional analysis relies on this assumption to find the exponents. It won't work for logarithmic or trigonometric relations, but for basic physical dependencies, it's a powerful tool.

Q: How did we get the dimensions of Intensity?

Intensity is defined as Power per unit Area. Power is Work done per unit time ($ML^2T^{-3}$). Area is $L^2$. Dividing the two gives $MT^{-3}$.

Q: Why did we ignore the constant $k$?

Dimensional analysis cannot determine dimensionless constants (like $2\pi$ or $1/2$). Since $k$ has no dimensions, it doesn't affect the balance of $M$, $L$, and $T$ on either side of the equation, so we can safely ignore it while finding the exponents.

Q: Do we need to solve for $c$ (the power of frequency)?

Not in this specific question! The problem only asks for the relationship between distance $d$ and intensity $S$. Once we found $b=1/3$, we had our answer. Solving for $c$ would be extra work, though it's good practice!

Q: What if the dimensions on both sides don't match?

If the dimensions don't match, it means the assumed physical relationship is incorrect, or a crucial physical quantity is missing from the assumption. The Principle of Homogeneity is a strict rule: an equation must be dimensionally balanced to be physically valid.

Question

To find the distance $d$ over which a signal can be seen clearly in foggy conditions, a railway 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

Sigma Percentile · Accepted Answer

The Foggy Signal: A Masterclass in Dimensional Analysis

Imagine you are a railway engineer. A thick, impenetrable fog has rolled in, and you need to calculate exactly how far a signal light can be seen. You know the distance must depend on the density of the fog, the intensity of the light, and the frequency of the light waves. But how do these variables connect?

This is where the magic of **Dimensional Analysis** comes in. Even without knowing the exact complex physics equations, we can deduce the relationship between these quantities simply by ensuring their fundamental units balance out. Let's dive into this elegant problem.

### Setting Up the Master Equation

We start by assuming a general proportional relationship. Let the distance $d$ depend on the fog's mass density $\rho$, the light's intensity $S$, and its frequency $f$. We can write this mathematically as:

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

Here, $k$ is a dimensionless constant, and $a$, $b$, and $c$ are the unknown powers we need to find. Our primary goal is to find the value of $b$, since the problem asks for the relationship between distance and intensity.

### Breaking Down the Dimensions

To use the Principle of Dimensional Homogeneity, we must express every quantity in terms of the fundamental dimensions: Mass ($M$), Length ($L$), and Time ($T$).

1. **Distance ($d$)**: This is simply a length.
   $$[d] = [L]$$

2. **Density ($\rho$)**: Mass per unit volume.
   $$[\rho] = \frac{[M]}{[L^3]} = [ML^{-3}]$$

3. **Intensity ($S$)**: The problem helpfully defines this as power per unit area. Power is work done per unit time ($ML^2T^{-3}$), and area is $L^2$.
   $$[S] = \frac{[ML^2T^{-3}]}{[L^2]} = [MT^{-3}]$$

4. **Frequency ($f$)**: Cycles per unit time.
   $$[f] = [T^{-1}]$$

### The Principle of Homogeneity

The Principle of Homogeneity states that for any valid physical equation, the dimensions on the left-hand side must perfectly match the dimensions on the right-hand side. Let's substitute our dimensional formulas into the master equation:

$$[L] = [ML^{-3}]^a \cdot [MT^{-3}]^b \cdot [T^{-1}]^c$$

Now, we group the bases together on the right side:

$$[M^0 L^1 T^0] = [M^{a+b} \cdot L^{-3a} \cdot T^{-3b-c}]$$

### Solving the Puzzle

We now have a system of linear equations by equating the exponents of $M$, $L$, and $T$ from both sides.

For Mass ($M$):
$$0 = a + b$$

For Length ($L$):
$$1 = -3a$$

From the length equation, we can immediately solve for $a$:
$$a = -\frac{1}{3}$$

Substitute this value of $a$ into the mass equation:
$$0 = -\frac{1}{3} + b \implies b = \frac{1}{3}$$

Notice that we don't even need to set up the equation for Time ($T$) or solve for $c$! The problem only asks for the dependency of distance on intensity, which is governed by the exponent $b$.

### The Final Conclusion

We have found that $b = \frac{1}{3}$. Substituting this back into our proportional relationship, we get:

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

The problem states that the engineer finds $d$ is proportional to $S^{1/n}$. By directly comparing our result with the given expression, the answer becomes crystal clear:

**$$n = 3$$**

And there we have it! By simply balancing the fundamental units of the universe, we've solved a complex engineering problem. Dimensional analysis is truly a superpower in physics.

Enter Numerical Value:

Visualized Solution

$d \propto ρ^{a} S^{b} f^{c}$

$Dimensions of Quantities$

$Principle of Homogeneity$

$Equating Powers of M and L$

$Solving for a and b$

$Finding n$

The Sigma Insight: Dimensional Analysis

Setting Up the Master Equation

Breaking Down the Dimensions

The Principle of Homogeneity

Solving the Puzzle

The Final Conclusion

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d∝ρaSbfc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensions of Quantities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Homogeneity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Powers of M and L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for a and b

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding n