Question 80: Waves Optics

Question

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

Accepted Answer

In Young's double-slit interference experiment, the width of one slit is half that of the other. The intensity, $ \mathrm{I} $, is proportional to the slit width. This gives us:
$ \begin{align*} \mathrm{I}_1 &= \mathrm{I}_0, \ \mathrm{I}_2 &= 2 \mathrm{I}_0. \end{align*} $
The maximum intensity, $ \mathrm{I}_{\max} $, occurs when the amplitudes add constructively:
$ \mathrm{I}_{\max} = \left(\sqrt{\mathrm{I}_1} + \sqrt{\mathrm{I}_2}ight)^2. $
The minimum intensity, $ \mathrm{I}_{\min} $, occurs when the amplitudes interfere destructively:
$ \mathrm{I}_{\min} = \left(\sqrt{\mathrm{I}_1} - \sqrt{\mathrm{I}_2}ight)^2. $
Now, using $ \mathrm{I}_1 = \mathrm{I}_0 $ and $ \mathrm{I}_2 = 2 \mathrm{I}_0 $, we find:
$ \frac{\mathrm{I}_{\max}}{\mathrm{I}_{\min}} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}. $

Answer

$3: 1$

Answer

$(2 \sqrt{2}+1):(2 \sqrt{2}-1)$

Answer

$9: 1$

Answer

$(3+2 \sqrt{2}):(3-2 \sqrt{2})$

Waves Optics

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

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