Question:
In a double slit experiment instead of taking slits of equal
widths, one slit is made twice as wide as the other, then in the
interference pattern
In a double slit experiment instead of taking slits of equal
widths, one slit is made twice as wide as the other, then in the
interference pattern
Updated On: May 19, 2022
- the intensities of both the maxima and the minima increases
- the intensity of the maxima increases and the minima has zero intensity
- the intensity of maxima decreases and that of minima increases
- the intensity of maxima decreases and the minima has zero intensity
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The Correct Option is A
Solution and Explanation
In interference we know that
$\hspace20mm I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2$
and $\hspace20mm I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$
Under normal conditions (when the widths of both the slits
are equal)
$\hspace20mm I_1=I_2=I$ (say)
$\therefore \hspace20mm I_{max}=4I \, \, and \, \, I_{min}=0$
When the width of one of the slits is increased. Intensity due
to that slit would increase, while that of the other will remain
same. So, let:
$\hspace25mm I_1=I$
and $\hspace25mm I_2=\eta I \hspace10mm (\eta > 1)$
Then, $\hspace20mm I_{max}=I(1+\sqrt {\eta})^2 > 4I$
and $\hspace20mm I_{min}=I(\sqrt {\eta}-1)^2 > 0$
$\therefore$ Intensity of both maxima and minima is increased.
$\hspace20mm I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2$
and $\hspace20mm I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$
Under normal conditions (when the widths of both the slits
are equal)
$\hspace20mm I_1=I_2=I$ (say)
$\therefore \hspace20mm I_{max}=4I \, \, and \, \, I_{min}=0$
When the width of one of the slits is increased. Intensity due
to that slit would increase, while that of the other will remain
same. So, let:
$\hspace25mm I_1=I$
and $\hspace25mm I_2=\eta I \hspace10mm (\eta > 1)$
Then, $\hspace20mm I_{max}=I(1+\sqrt {\eta})^2 > 4I$
and $\hspace20mm I_{min}=I(\sqrt {\eta}-1)^2 > 0$
$\therefore$ Intensity of both maxima and minima is increased.
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Concepts Used:
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- Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
- The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
- Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
Read More: Young’s Double Slit Experiment