Question

If the ratio of the intensity of wave from two coherent sources is 4, then the visibility of the fringes is

• A. $9$
• B. $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$$}\right.5$
• C. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• D. $4$

Answer 1

The correct option is B

$\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$$}\right.5$

Step 1. Diven data:

Let the intensities be $I_1$ and $I_2$

Given $\frac{I_1}{I_2}$=$4$, so $I_1=4I_2$

Step 2. Formula used:

$I_{max}={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$

$I_{min}={\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2$

Step 3. Explanation:

We know $I_{max}={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2={\left(2\sqrt{I_2}+\sqrt{I_2}\right)}^2={\left(3\sqrt{I_2}\right)}^2=9I_2$

Also $I_{min}={\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2={\left(2\sqrt{I_2}-\sqrt{I_2}\right)}^2={\left(\sqrt{I_2}\right)}^2=I_2$

Now $$\begin{gathered} I_{max}-I_{min}=9I_2-I_2=8I_2 \\ {I}_{max}+I_{min}=9I_2+I_2=10I_2 \\ \Rightarrow \left(\frac{I_{max}-I_{min}}{I_{max}+I_{min}}\right)=\left(\frac{8I_2}{10I_2}\right)=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right. \end{gathered}$$

Hence the correct option is option B.