91Ó°ÊÓ

The strength or amount of light produced by a certain lighting source is referred to as light intensity. It is a power measurement of a light source that is wavelength-weighted..

The following equation gives the total double-slit intensity:

$I_{\text{double}}=I_0{\left[\frac{\sin (\mathrm{π}ay/\mathrm{λ}L)}{\mathrm{π}ay/\mathrm{λ}L}\right]}^2{\cos }^2(\mathrm{π}dy/\mathrm{λ}L)$

We must first locate the midway point between the center and the first minimum in order to find $I_{\text{double}}$as a function $I_o$,To do so, we can use the following equation to explain the position of dark fringes:

$y_m^{\text{'}}=\left(m+\frac{1}{2}\right)\frac{\mathrm{λ}L}{d}m=0,1,2,…$

The first minimum's position is

$y_0^{\text{'}}=\frac{\mathrm{λ}L}{2d}$

And the halfway point is at half of this number, implying that

$y_{\text{half}}=\frac{y_0}{2}=\frac{\mathrm{λ}L}{4d}$

Substitute the values,

$I_{\text{double}}=I_0{\left(\frac{\sin \left(\mathrm{π}a\left(\frac{\mathrm{λ}L}{4d}\right)/\mathrm{λ}L\right)}{\mathrm{π}a\left(\frac{\mathrm{λ}L}{4d}\right)/\mathrm{λ}L}\right)}^2{\cos }^2\left(\mathrm{π}d\left(\frac{\mathrm{λ}L}{4d}\right)/\mathrm{λ}L\right)$

$=I_0{\left(\frac{\sin \left(\frac{\mathrm{Ï€}a}{4d}\right)}{\frac{\mathrm{Ï€}a}{4d}}\right)}^2{\cos }^2\left(\frac{\mathrm{Ï€}}{4}\right)$

$=I_0{\left(\frac{\sin \left(\frac{\mathrm{π}×0.12\mathrm{mm}}{4×0.3\mathrm{mm}}\right)}{\frac{\mathrm{π}×0.12\mathrm{mm}}{4×0.3\mathrm{mm}}}\right)}^2{\cos }^2\left(\frac{\mathrm{π}}{4}\right)$

$I_{\text{double}}=0.48I_0$