Interference Pattern Amplitude Ratio Analysis

This question involves understanding the relationship between the amplitudes of coherent light sources and the intensities observed in an interference pattern. We need to find a specific ratio related to the maximum and minimum intensities based on the given ratio of amplitudes.

Understanding Light Interference and Intensity

When two coherent light sources interfere, they produce an interference pattern characterized by alternating bright and dark fringes. The brightness of these fringes depends on the intensities of the light waves, which in turn depend on their amplitudes.

• Coherent Sources: Light sources that maintain a constant phase difference between them.
• Amplitude (a): The maximum displacement or magnitude of the oscillation of the light wave.
• Intensity (I): The power per unit area carried by the light wave. It is directly proportional to the square of the amplitude. Mathematically, $I \propto a^2$. We can write this as $I = k a^2$, where $k$ is a constant of proportionality.

Calculating Maximum and Minimum Intensities

Let the amplitudes of the two coherent light sources be $A_1$ and $A_2$. The problem states that the ratio of their amplitudes is $n$. We can write this as:

$ \frac{A_1}{A_2} = n $

Without loss of generality, let's assume $A_1 = n A_2$.

In an interference pattern:

• The maximum intensity ($I_{max}$) occurs when the waves interfere constructively (crests align with crests). The resultant amplitude is the sum of the individual amplitudes, $A_1 + A_2$.

$ I_{max} \propto (A_1 + A_2)^2 $

• The minimum intensity ($I_{min}$) occurs when the waves interfere destructively (crests align with troughs). The resultant amplitude is the difference between the individual amplitudes, $|A_1 - A_2|$.

$ I_{min} \propto (A_1 - A_2)^2 $

Using the proportionality $I = k a^2$, we have:

$ I_{max} = k (A_1 + A_2)^2 $

$ I_{min} = k (A_1 - A_2)^2 $

Deriving the Required Ratio

We need to calculate the ratio $\frac{{{I_{max}} - {I_{min}}}}{{{I_{max}} + {I_{min}}}}$. Let's substitute the expressions for $I_{max}$ and $I_{min}$:

$ \frac{{{I_{max}} - {I_{min}}}}{{{I_{max}} + {I_{min}}}} = \frac{k (A_1 + A_2)^2 - k (A_1 - A_2)^2}{k (A_1 + A_2)^2 + k (A_1 - A_2)^2} $

The constant $k$ cancels out:

$ \frac{{(A_1 + A_2)^2 - (A_1 - A_2)^2}}{{(A_1 + A_2)^2 + (A_1 - A_2)^2}} $

Now, let's use the given amplitude ratio $\frac{A_1}{A_2} = n$. Substitute $A_1 = n A_2$ into the expression:

$ \frac{{(n A_2 + A_2)^2 - (n A_2 - A_2)^2}}{{(n A_2 + A_2)^2 + (n A_2 - A_2)^2}} $

Factor out $A_2$ from each term:

$ \frac{{(A_2(n + 1))^2 - (A_2(n - 1))^2}}{{(A_2(n + 1))^2 + (A_2(n - 1))^2}} $

$ \frac{{A_2^2(n + 1)^2 - A_2^2(n - 1)^2}}{{A_2^2(n + 1)^2 + A_2^2(n - 1)^2}} $

The $A_2^2$ term cancels out from the numerator and the denominator:

$ \frac{{(n + 1)^2 - (n - 1)^2}}{{(n + 1)^2 + (n - 1)^2}} $

Let's expand the squares:

• Numerator: $(n + 1)^2 - (n - 1)^2 = (n^2 + 2n + 1) - (n^2 - 2n + 1) = n^2 + 2n + 1 - n^2 + 2n - 1 = 4n$.
• Denominator: $(n + 1)^2 + (n - 1)^2 = (n^2 + 2n + 1) + (n^2 - 2n + 1) = n^2 + 2n + 1 + n^2 - 2n + 1 = 2n^2 + 2 = 2(n^2 + 1)$.

Substituting these back into the ratio:

$ \frac{{4n}}{{2(n^2 + 1)}} $

Simplifying this expression gives:

$ \frac{{2n}}{{{n^2} + 1}} $

Conclusion

The calculated ratio $\frac{{{I_{max}} - {I_{min}}}}{{{I_{max}} + {I_{min}}}}$ based on the amplitude ratio $n$ is $\frac{{2n}}{{{n^2} + 1}}$. This matches one of the provided options.

Final Answer Check

Comparing our result with the options:

1. $\frac{{{n^2} - 1}}{{{n^2} + 1}}$
2. $\frac{{2\sqrt n }}{{n + 1}}$
3. $\frac{{2n}}{{{n^2} + 1}}$
4. $\frac{n}{{{n^2} + 1}}$
5. (No expression provided)

Our derived ratio $\frac{{2n}}{{{n^2} + 1}}$ corresponds exactly to option 3.