Question 268: Wave Optics

Question

In Young's double slit interference experiment, using two coherent sources of different amplitudes, the intensity ratio between bright to dark fringes is $5: 1$. The value of the ratio of resultant amplitudes of bright fringe to dark fringe is

Accepted Answer

Given:
In Young's double slit experiment, two coherent sources have different amplitudes.
The intensity ratio between
bright
and
dark
fringes is $5:1$.
Let the amplitudes of the two sources be $a_1$ and $a_2$.
Step 1: Write the formulas
Maximum intensity (bright fringe):
$ I_{	ext{max}} = (a_1 + a_2)^2 $
Minimum intensity (dark fringe):
$ I_{	ext{min}} = (a_1 - a_2)^2 $
Step 2: Use the given intensity ratio
$ \frac{I_{	ext{max}}}{I_{	ext{min}}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{5}{1} $
Step 3: Take square root on both sides
$ \frac{a_1 + a_2}{a_1 - a_2} = \sqrt{5} $
Step 4: Find the required ratio
The
ratio of amplitudes at bright fringe to dark fringe
is:
$ \frac{A_	ext{bright}}{A_	ext{dark}} = \frac{a_1 + a_2}{|a_1 - a_2|} $
But from the above,
$ \frac{a_1 + a_2}{a_1 - a_2} = \sqrt{5} $
So,
$ \frac{A_	ext{bright}}{A_	ext{dark}} = \sqrt{5} : 1 $
Final Answer
$ \boxed{	ext{Option B: } \sqrt{5} : 1} $

Answer

$\left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)$

Answer

$\sqrt{5}: 1$

Answer

$\left(\frac{\sqrt{5}-1}{\sqrt{5}+1}\right)$

Answer

$1: \sqrt{5}$

Wave Optics

In Young's double slit interference experiment, using two coherent sources of different amplitudes, the intensity ratio between bright to dark fringes is 5: 1. The value of the ratio of resultant amplitudes of bright fringe to dark fringe is

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