Question

$FIGUREP33.36$shows the light intensity on a screen behind a double slit. Suppose one slit is covered. What will be the light intensity at the center of the screen due to the remaining slit?

Short answer

At the remaining slit, the sunshine intensity at the center of the screen is $I_0=3\mathrm{mW}/{\mathrm{m}}^2.$

Step-by-step solution

Step: 1 Intensity of fringes:

There are two types of fringes that form in an interference pattern: brilliant fringes and dark fringes.The fringe intensity changes betting on the space from the screen, the space between the slits, and the phase point.

Step: 2 Equating the equation:

It's essential to understand that the intensity of the the perimetter pattern produced by a double-slit diffraction experiment is expressed as in terms of the intensity I 0 of the sunshine originating from one slit.

$I=4I_0\cos \left(\frac{\pi d}{\lambda L}y\right).$

Step: 3 Finding the light intensity:

This indicates that the intensity produced by a single slit, which is what we're interested in, will be one-fourth of the highest intensity seen in the double-slit experiment. Given that the latter is $12mW/m^2$, if a slit is blocked, the intensity at the centre of the screen will be

$$\begin{gathered} I=4I_0\cos \left(\frac{\pi d}{\lambda L}y\right) \\ {I}_0=3\mathrm{mW}/{\mathrm{m}}^2. \end{gathered}$$