Question

In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is $3 .$ Then, the value of the ratio of the amplitudes of the wave that arrive there is:

• A. $\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$
• B. $\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$
• C. $\sqrt{3}: 1$
• D. $1: \sqrt{3}$

Answer 1

Answer: (A), (B)

According to given question,
$I_{1}: I_{2}=3: 1, a_{1}: a_{2}=$?
The intensity of light due to an illuminated slit is directly proportional to the width of the slit, if $w_{1}$ and $w_{2}$ be the width of two silts and $I_{1}$ and $I_{2}$ are the intensities of light on the screen due to the respective slits then
$\frac{w_{1}}{w_{2}}=\frac{I_{1}}{I_{2}}=\left(\frac{a_{1}}{a_{2}}\right)^{2}$
where $a_1$, and $a_2$, are the amplitudes of the corresponding light wave
Here, $\frac{w_{1}}{w_{2}}=\frac{3}{1}$
$ \Rightarrow \left(\frac{a_{1}}{a_{2}}\right)^{2}=\frac{3}{1}$ or $a_{1}: a_{2}=\sqrt{3}: 1$