Answer
$0.242$
Work Step by Step
The intensity of the diffraction pattern at any given point angle $\theta$ is
$I(\theta)=I_m\Big(\frac{\sin\alpha}{\alpha}\Big)^2$
where $I_m$ is the intensity at the center of the pattern and
$\alpha=\frac{\pi}{\lambda}a\sin\theta$
In this problem, the wavelength of the light is $\lambda=500\;nm$, the slit width is $a=6.00\;\mu m$, and the viewing screen is at distance $D=3.00\;m$. The intensity of the diffracted light at point P at $y=15.0\;cm$ is represented by $I_p$.
As, $\theta$ is the location of P with respect to the central axis, therefore
$\sinθ=\frac{y}{\sqrt {D^2+y^2}}$
$\therefore\;\;\alpha=\frac{\pi a}{\lambda}\times\frac{y}{\sqrt {D^2+y^2}}$
$\implies\;\;\alpha=\frac{\pi\times 6\times 10^{-6}}{500\times 10^{-9}}\times\frac{0.15}{\sqrt {3^2+0.15^2}}$
$\implies\;\;\alpha\approx1.883\;rad$
Therefore,
$\frac{I_p}{I_m}=\Big(\frac{\sin\alpha}{\alpha}\Big)^2=\Big\{\frac{\sin(1.883)}{1.883}\Big\}^2\approx0.242$
