Question

To find the distance $ d $ over which a signal can be seen clearly in a 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^{\dfrac{1}{n}}} $ . Find value of $ n $ .

Answer 1

Hint: Mass density is directly proportional to mass and inversely proportional to the volume, while the frequency is inversely proportional to time. Power is inversely proportional to time but directly proportional to force and length.

Formula used: $ d = L $ where $ d $ is distance and $ L $ is the dimension of length, $ \rho = M{L^{ - 3}} $ where $ M $ is the dimension of mass and $ \rho $ is the mass density. $ S(joule/\sec )/area = M{T^{ - 3}} $ where $ T $ is the dimension of time.

Complete step by step answer
The dimensions of each variable are
 $ \Rightarrow d = L $ (dimension of length)
 $ \Rightarrow \rho = M{L^{ - 3}} $ (dimension of mass per unit volume)
 $ \Rightarrow S(joule/\sec )/area)) = M{T^{ - 3}} $
 $ \Rightarrow f = {T^{ - 1}} $
To analyze using dimensions we say
 $ \Rightarrow d = {(S)^x}{(\rho )^y}{(f)^z} $
Replacing them with their dimensions, it becomes
 $ \Rightarrow L = {(M{T^{ - 2}})^x}{(M{L^{ - 3}})^y}{({T^{ - 1}})^z} $
 $ \Rightarrow L = {M^x}{T^{ - 2x}}({M^y}{L^{ - 3y}})({T^{ - z}}) $
Using the rules of indices,
 $ \Rightarrow L = {M^{x + y}}{L^{ - 3y}}{T^{ - 2x - z}} $
Equating powers of equal base
 $ \Rightarrow - 3y = 1;x + y = 0; - 2x - z = 0 $
 $ \Rightarrow y = - \dfrac{1}{3};x = - y;z = - 2x $
Therefore,
 $ \Rightarrow x = \dfrac{1}{3} $
Which is the only variable that matters, since intensity was raised to the power of $ x $ .
Thus, $ {S^x} = {S^{\dfrac{1}{3}}} $ .
Since $ S $ is raised to the power of $ n $ as given in the question, we get
 $ \Rightarrow {S^{\dfrac{1}{n}}} = {S^{\dfrac{1}{3}}} $
 $ \Rightarrow n = 3 $
Using the law of exponents we get,
 $ \Rightarrow \dfrac{1}{n} = \dfrac{1}{3} $
 $ \Rightarrow n = 3 $
Hence, the value of $ n $ is 3.

Additional Information
Dimensional analysis can be a very powerful tool even in reality. The engineer above utilized it to derive the expressions of a quantity in terms of another. But other uses of dimensional analysis includes: To convert units from one system to another, to verify consistency of dimensional equations. However, its limitations include: blindness to dimensional constant, cannot be used for logarithmic or trigonometric functions.

Note
For proper understanding, let us derive the dimension of intensity from first principles.
Intensity is given by $ S = \dfrac{{Power}}{{Area}} $ ,
In turn power is given by $ P = \dfrac{{Energy}}{{time}} $
Also, energy is given by $ E = Force \times length $
In fundamental units, force has a unit of $ kgm/{s^2} $ which in dimensional analysis translates to $ ML{T^{ - 2}} $ .
Thus, energy has a dimension of $ ML{T^{ - 2}} \times L $ and power has a dimension of $ M{L^2}{T^{ - 2}}/T $ .
Thus, intensity has a dimension of $ M{L^2}{T^{ - 3}}/{L^2} = M{T^{ - 3}} $ .