Question

The intensity of the light coming from one of the slits in Young's experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern?​

Answer 1

Let ,

The intensity of the 1 st slit be I

The intensity of the 2 nd slit be I '

As the intensity of the light coming from one of the slits in Young's experiment is twice the intensity of the light coming from the other slit.

I '= 2× I

Bright fringes ( maximum) intensity

$I max= {( \sqrt{ I } + \sqrt{ I' } )}^{2}$

$I max = { ( \sqrt{ I } + \sqrt{2 I } )}^{2}$

Dark fringes ( minimum) intensity

$I min= {( \sqrt{ I } - \sqrt{ I' } )}^{2}$

$I min= {( \sqrt{ I } - \sqrt{ 2I } )}^{2}$

Ratio of the intensities of the bright and dark fringes,

$\frac{I max }{I min} = \frac{{ ( \sqrt{ I } + \sqrt{2 I } )}^{2} }{{ ( \sqrt{ I } - \sqrt{2 I } )}^{2} }$

$\frac{I max }{I min} = \frac{{ ( 1 + \sqrt{2 } )}^{2}I }{{ (1 - \sqrt{2 } )}^{2} I}$

$\frac{I max }{I min} = \frac{{ ( 1 + \sqrt{2 } )}^{2} }{{ (1 - \sqrt{2 } )}^{2} }$