Question

Ratio of maximum to minimum intensity is 25:16. Calculate ratio of intensities:
A. 5:4
B. 4:5
C. 9:1
D. 81:1

Answer 1

Hint: We require the basic idea of the phenomena of interference of light. If we know how intensity depends on the two amplitudes, we can use the given data to find the answer. We just require the expressions for the maximum and minimum intensities in terms of amplitudes.

Formula required: $\dfrac{I_{max}}{I_{min}}=(\dfrac{a_1+a_2}{a_1-a_2})^2$

Complete step by step solution:
First of all, interference occurs when two rays of light of the same frequency emit from two coherent sources and fall on a screen placed near them. The view that we get on the screen is an alternate pattern of bright and dark bands.
Let, $a_1$ and $a_2$ be the amplitudes of the two beams of light that will interfere $\left(a_1 > a_2\right)$. Now, if $\delta$ be the phase difference between these two beams of light, the resultant intensity is given be,
$I=a_1^2+a_2^2+2.a_1.a_2.\cos\delta$
So, for $\cos\delta=+1$, intensity is maximum, $I_{max}=(a_1+a_2)^2$
The lines with this intensity are called the bright fringe in the interference pattern. For this, the value of the phase difference $\delta$ must be an integral multiple of $2\phi$. That is $\delta=2n\phi$ where n is any integer.
Similarly, for $cos\delta=-1$, intensity is minimum, $I_{min}=(a_1-a_2)^2$
The lines with the minimum intensity are called the dark fringe in the interference pattern. For this, the value of the phase difference $\delta$ must be an odd integral multiple of $\phi$. That is $\delta=(2n+1)\phi$ where n is any integer.
Now, simply taking the ratio of the maximum intensity to the minimum intensity, we obtain,
$\dfrac{I_{max}}{I_{min}}=(\dfrac{a_1+a_2}{a_1-a_2})^2$
Putting, $\dfrac{I_{max}}{I_{min}}=\dfrac{25}{16}$, we obtain,
$\dfrac{a_1+a_2}{a_1-a_2}=\dfrac{5}{4}$
Then, $\dfrac{a_1+a_2+ a_1-a_2}{a_1+a_2- a_1+a_2}=\dfrac{5+4}{5-4}$
$\dfrac{a_1}{a_2}=\dfrac{9}{1}$
$\dfrac{I_1}{I_2}=\dfrac{a_1^2}{a_2^2}=\dfrac{81}{1}$
$I_1$ and $I_2$ are the intensities of the two beams.
So, option D is the correct answer.

Additional information: If x be the path difference between the two beams, then the phase difference is given by,
$\delta=\dfrac{2\phi}{\lambda}.x$
Where $\lambda$ is the wavelength of the light.

Note: Remember the following things,
1. Never take the negative values of the square root.
2. Write $(a_1-a_2)$ only if $a_1 > a_2$
3. The intensity is the square of amplitude only for individual beams. After interference, the formula changes.