Question

Two coherent light sources having intensities in the ratio $2\mathrm{x}$ produce an interference pattern. The ratio $\frac{({\mathrm{I}}_{\max }-{\mathrm{I}}_{\min })}{({\mathrm{I}}_{\max }+{\mathrm{I}}_{\min })}$ will be:

• A. $2\sqrt{\frac{\left(2x\right)}{x+1}}$
• B. $\sqrt{\frac{\left(2x\right)}{2x+1}}$
• C. $2\sqrt{\frac{2x}{(2x+1)}}$
• D. $\sqrt{\frac{\left(2x\right)}{x+1}}$

Answer 1

The correct option is C

$2\sqrt{\frac{2x}{(2x+1)}}$

Step-1: Given data

Let the intensities be ${\mathrm{I}}_1\mathrm{and}{\mathrm{I}}_2$ then it is given that $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}=2x$

Step-2: To find

$\frac{({\mathrm{I}}_{\max }-{\mathrm{I}}_{\min })}{({\mathrm{I}}_{\max }+{\mathrm{I}}_{\min })}$

Step-3: Calculation

For two coherent sources, the maximum and minimum intensity is given as:

$$\begin{gathered} {\mathrm{I}}_{\max }={\mathrm{I}}_1+{\mathrm{I}}_2+2\sqrt{{\mathrm{I}}_1{\mathrm{I}}_2}={(\sqrt{{\mathrm{I}}_1}+\sqrt{{\mathrm{I}}_2})}^2 \\ {\mathrm{I}}_{\min }={\mathrm{I}}_1+{\mathrm{I}}_2-2\sqrt{{\mathrm{I}}_1{\mathrm{I}}_2}={(\sqrt{{\mathrm{I}}_1}+\sqrt{{\mathrm{I}}_1})}^2 \end{gathered}$$

On rearranging

$$\begin{gathered} {\mathrm{I}}_{\max }={(\sqrt{{\mathrm{I}}_1}+\sqrt{{\mathrm{I}}_2})}^2 \\ {\mathrm{I}}_{\max }={\mathrm{I}}_2{(\sqrt{\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}}+1)}^2{\mathrm{I}}_{\max }={\mathrm{I}}_2{(\sqrt{2\mathrm{x}}+1)}^2 \end{gathered}$$

and,

$$\begin{gathered} {\mathrm{I}}_{\min }={(\sqrt{{\mathrm{I}}_1}-\sqrt{{\mathrm{I}}_2})}^2 \\ {\mathrm{I}}_{\min }={\mathrm{I}}_2{(\sqrt{\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}}-1)}^2{\mathrm{I}}_{\max }={\mathrm{I}}_2{(\sqrt{2\mathrm{x}}-1)}^2 \end{gathered}$$

Adding maximum and minimum value

$$\begin{gathered} {\mathrm{I}}_{\max }+{\mathrm{I}}_{\min }={\mathrm{I}}_2{(\sqrt{2\mathrm{x}}+1)}^2+{\mathrm{I}}_2{(\sqrt{2\mathrm{x}}-1)}^2 \\ \Rightarrow {\mathrm{I}}_{\max }+{\mathrm{I}}_{\min }={\mathrm{I}}_2\left[{(\sqrt{2\mathrm{x}}+1)}^2+{(\sqrt{2\mathrm{x}}-1)}^2\right]------\left(\mathrm{i}\right) \end{gathered}$$

Substracting maximum and minimum value

$$\begin{gathered} {\mathrm{I}}_{\max }-{\mathrm{I}}_{\min }={\mathrm{I}}_2{(\sqrt{2\mathrm{x}}+1)}^2-{\mathrm{I}}_2{(\sqrt{2\mathrm{x}}-1)}^2 \\ \Rightarrow {\mathrm{I}}_{\max }-{\mathrm{I}}_{\min }={\mathrm{I}}_2\left[{(\sqrt{2\mathrm{x}}+1)}^2-{(\sqrt{2\mathrm{x}}-1)}^2\right]------\left(\mathrm{ii}\right) \end{gathered}$$

Now, Dividing eq(ii) by eq(i)

$\frac{{\mathrm{I}}_{\max }-{\mathrm{I}}_{\min }}{{\mathrm{I}}_{\max }+{\mathrm{I}}_{\min }}=\frac{{\mathrm{I}}_2\left[{(\sqrt{2\mathrm{x}}+1)}^2-{(\sqrt{2\mathrm{x}}-1)}^2\right]}{{\mathrm{I}}_2\left[{(\sqrt{2\mathrm{x}}+1)}^2+{(\sqrt{2\mathrm{x}}-1)}^2\right]}$

Rationalizing the denominator, we get

$$\begin{gathered} \frac{{\mathrm{I}}_{\max }-{\mathrm{I}}_{\min }}{{\mathrm{I}}_{\max }+{\mathrm{I}}_{\min }}=4\sqrt{\frac{\left(2x\right)}{(2+4x)}} \\ \Rightarrow \frac{{\mathrm{I}}_{\max }-{\mathrm{I}}_{\min }}{{\mathrm{I}}_{\max }+{\mathrm{I}}_{\min }}=2\sqrt{\frac{\left(2x\right)}{(2x+1)}} \end{gathered}$$

Hence, Option(C) is correct.