Question

If one of two identical slits producing interference in young's double slit experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.

Answer 1

Let initial intensity of light passing through each slit is I₀ .
According to question,
one of two slits is covered with glass as shown in figure ,such that the intensity of light passing through it is reduced to 50% .
Means intensity of this slit will be I₀/2 after covering with glass .
Now, use formula of intensity ,
$\bold{I=I_1+I_2+2\sqrt{I_1I_2}cos\Phi}$
Here I₁ , I₂ are the intensities of both slits respectively ,
Φ is the phase difference

For Maximum intensity , Φ = 0°
Then, $\bold{I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2}\\\\\text{and,}\:\bold{I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2}$[at Ф = π]
Here, I₁ = I₀ and I₂ = I₀/2

$\bold{I_{max}=[\sqrt{I_0}+\sqrt{\frac{I_0}{2}}]^2}$
And $\bold{I_{min}=[\sqrt{I_0}-\sqrt{\frac{I_0}{2}}]^2}$
Now, ratio of Imax and Imin ,
$\bold{\frac{I_{max}}{I_{min}}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2}}$