Question 257: Waves Optics

Question

The two coherent sources produce interference with intensity ratio ' $b$ '. In the interference pattern, the ratio $\frac{I_{	ext {max }}+I_{	ext {min }}}{I_{	ext {max }}-I_{	ext {min }}}$ will be

Accepted Answer

WeÃ¢ÂÂre asked:
Two coherent sources interfere. Intensity ratio = $b$. Then compute
$ \frac{I_{\max}+I_{\min}}{I_{\max}-I_{\min}}\,. $
Step 1. Express intensities of sources
Let the two sources have intensities $I_1$ and $I_2$.
Their ratio:
$ \frac{I_1}{I_2} = b. $
Assume $I_1 = b I_2$.
Let $I_2 = I_0$. Then $I_1 = b I_0$.
Step 2. Amplitudes
Amplitude is proportional to $\sqrt{I}$.
So,
$ A_1 = \sqrt{I_1} = \sqrt{b I_0} = \sqrt{b}\,\sqrt{I_0}, $
$ A_2 = \sqrt{I_2} = \sqrt{I_0}. $
Step 3. Constructive and destructive interference
Maximum amplitude: $A_{\max} = A_1 + A_2 = (\sqrt{b}+1)\sqrt{I_0}$.
So maximum intensity:
$ I_{\max} = (A_{\max})^2 = (\sqrt{b}+1)^2 I_0. $
Minimum amplitude: $A_{\min} = |A_1 - A_2| = |\sqrt{b}-1|\sqrt{I_0}$.
So minimum intensity:
$ I_{\min} = (A_{\min})^2 = (\sqrt{b}-1)^2 I_0. $
Step 4. Compute the required ratio
$ \frac{I_{\max} + I_{\min}}{I_{\max} - I_{\min}}  = \frac{(\sqrt{b}+1)^2 I_0 + (\sqrt{b}-1)^2 I_0}{(\sqrt{b}+1)^2 I_0 -(\sqrt{b}-1)^2 I_0}. $
Cancel $I_0$:
$ = \frac{(\sqrt{b}+1)^2 + (\sqrt{b}-1)^2}{(\sqrt{b}+1)^2 - (\sqrt{b}-1)^2}. $
Expand:
Numerator: $(b+1+2\sqrt{b}) + (b+1-2\sqrt{b}) = 2b+2.$
Denominator: $(b+1+2\sqrt{b}) - (b+1-2\sqrt{b}) = 4\sqrt{b}.$
So ratio:
$ = \frac{2(b+1)}{4\sqrt{b}} = \frac{1+b}{2\sqrt{b}}. $
Ã¢ÂÂ Final Answer:
Option
B
:
$ \boxed{\frac{1+b}{2\sqrt{b}}} $

Answer

$\frac{1+b}{\sqrt{b}}$

Answer

$\frac{1+b}{2 \sqrt{b}}$

Answer

$\frac{2 \sqrt{b}}{1+b}$

Answer

$\frac{2 \sqrt{b}}{(1+b)^2}$

Waves Optics

The two coherent sources produce interference with intensity ratio ' b '. In the interference pattern, the ratio \frac{I_{\text {max }}+I_{\text {min }}}{I_{\text {max }}-I_{\text {min }}} will be

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