Question 17: Waves Optics

Question

The interference pattern is obtained with two coherent light sources of intensity ratio $9: 1$.
The ratio of $\frac{I_{\operatorname{MAX}}+I_{\operatorname{MIN}}}{I_{\operatorname{MAX}}-I_{\operatorname{MIN}}}$ is $\frac{\alpha}{\beta}$. The values of $\alpha$ and $\beta$ are:

Accepted Answer

Let the two intensities be
$$I_1=9I_0,\quad I_2=1I_0$$
so the amplitudes are
$$a_1=\sqrt{9I_0}=3\sqrt{I_0},\quad a_2=\sqrt{I_0}.$$
Maximum intensity (inÃ¢ÂÂphase):
$$I_{\rm max}=(a_1+a_2)^2=(3\sqrt{I_0}+\sqrt{I_0})^2=16\,I_0.$$
Minimum intensity (outÃ¢ÂÂofÃ¢ÂÂphase):
$$I_{\rm min}=(a_1-a_2)^2=(3\sqrt{I_0}-\sqrt{I_0})^2=4\,I_0.$$
The desired ratio is
$$\frac{I_{\rm max}+I_{\rm min}}{I_{\rm max}-I_{\rm min}}      =\frac{16I_0+4I_0}{16I_0-4I_0}      =\frac{20}{12}      =\frac{5}{3}.$$
So $\alpha=5$ and $\beta=3$. (Option A)

Answer

5 and 3

Answer

3 and 1

Answer

1 and 9

Answer

9 and 1

Waves Optics

The interference pattern is obtained with two coherent light sources of intensity ratio 9: 1. The ratio of \frac{I_{\operatorname{MAX}}+I_{\operatorname{MIN}}}{I_{\operatorname{MAX}}-I_{\operatorname{MIN}}} is \frac{\alpha}{\beta}. The values of \alpha and \beta are:

Choose the correct answer: