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In a Young’s double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

Option 1 -

1:4

Option 2 -

4: 1

Option 3 -

3:1

Option 4 -

2:1

2 Views | Posted 2 months ago

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Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

Amplitude is proportional to the slit – width, thus,

^{$\frac{A_{1}}{A_{2}} = 3$}

^{$\frac{l_{m a x}}{l_{m i n}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}} = \frac{{(\frac{A_{1}}{A_{2}} + 1)}^{2}}{{(\frac{A_{1}}{A_{2}} - 1)}^{2}} = \frac{{(3 + 1)}^{2}}{{(3 - 1)}^{2}} = 4 .$}

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In a Young’s double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

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