Question

In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other, then, in the interference pattern

• A. the intensities of both maxima and minima increase
  
• B. the intensity of the maxima increases and the minima have zero intensity
  
• C. the intensity of the maxima decreases and that of the minima increases
  
• D. the intensity of the maxima decreases and the minima have zero intensity.

Answer 1

The correct option is A the intensities of both maxima and minima increase


In the case when the slits are equal width, the intensity of light emerging from the two slits is the same, say, $I_0.$ Then
$I_{max}=I_0+I_0+2\sqrt{I_0{I}_0}=4I_{0}and{I}_{min}=I_0+I_0-2\sqrt{I_0{I}_0}=0$
When one slit say $S_1$ is made twice as wide as the other, the intensity of light from $S_1$ is doubled, i.e. $I_1=2I_{0}but{I}_2=I_0.$ Hence, in this case
$I_{max}^{\prime }=I_1+I_2+2\sqrt{I_1{I}_2}=2I_0+I_0+2\sqrt{2I_0{I}_0}=3I_0+2\sqrt{2}{I}_0=5.83T_{0}and{I}_{min}^{\prime }=I_1+I_2-2\sqrt{I_1{I}_2}=2I_0+I_0-2\sqrt{2I_0{I}_0}=3I_0-2\sqrt{2}{I}_0=0.17I_{0}Thus{I}_{max}^{\prime }>I_{max}and{I}_{min}^{\prime }>I_{min},whichischoiceis\left(a\right).$