Question

(a) If one of two identical slits producing interference in Young's 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.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light ?

Answer 1

(a) The resultant intensity in Young's experiment is given by
$I_R=I_1+I_2+2\sqrt{I_1{I}_2}cos\varphi$

When slit is not covered, then $I_0$ is the intensity from each slit.

Maximum intensity $\left(I_{max}\right)$ occurs when $\Phi =0^{\circ }$
Minimum intensity $\left(I_{min}\right)$ occurs when $\Phi ={180}^{\circ }$

If one slit is covered with glass to reduce its intensity by 50%, then

$I_{max}=I_0+\frac{I_0}{2}+2\sqrt{I_0\times \frac{I_0}{2}}cos{0}^{\circ }$

= $1.5I_0+2\times 0.707I_0=2.914I_0$

$I_{min}=I_0+\frac{I_0}{2}+2\sqrt{I_0\times \frac{I_0}{2}}cos{180}^0$

$=1.5I_0-2\times 0.707I_0=0.086I_0$

Ratio = $\frac{I_{max}}{I_{min}}=\frac{{2.9141}_0}{{0.0861}_0}=33.884$

(b) If instead of monochromatic light, white light is used, then the central fringe will be white and the fringes on either side will be coloured. Blue colour will be nearer to central fringe and red will be farther away. The path difference at the centre on perpendicular bisector of slits will be zero for all colours and each colour produces a bright fringe thus resulting in white fringe. Further, the shortest visible wave, blue, produces a bright fringe first.