A laser beam with a wavelength of $480nm$ illuminates two $0.12-mm$-wide slits separated by $0.30mm$. The interference pattern is observed on a screen $2.3m$ behind the slits. What is the light intensity, as a fraction of the maximum intensity $I_0$, at a point halfway between the center and the first minimum?

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:

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

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:

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

The first minimum's position is

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

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

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

Substitute the values,

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

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

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

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