First slide

Young's double slit Experiment

Question

In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

Moderate

A

the intensities at the screen due to the two slits are 5 units and 4 units respectively
B

the intensities at the scree[ due to the two slits are 4 units and 1 unit respectively
C

the amplitude ratio is 3
D

the amplitude ratio is 4

Solution

$\frac{I_{max}}{I_{min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} Given that \frac{I_{max}}{I_{min}} = 9 \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} = 9 or \frac{(a_{1} + a_{2})}{(a_{1} - a_{2})} = 3 or a_{1} + a_{2} = 3 a_{1} - 3 a_{2} or a_{1} = 2 a_{2} Hence answer [d) is correct . Let I_{1} and I_{2} be the intensities due to two slits. Then I_{1} \propto {a_{1}}^{2} and I_{2} \propto {a_{2}}^{2} \frac{I_{1}}{I_{2}} = \frac{{a_{1}}^{2}}{{a_{2}}^{2}} = \frac{{(2 a_{2})}^{2}}{{a_{2}}^{2}} = \frac{4}{1} Hence aaswer (b) is correct$

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