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 $a_1:a_2=\sqrt{3}:1$