91影视

$位=580nm=580脳{10}^{-9}m.{胃}_1=卤{90}^{鈭�},胃=45.0^{鈭�}$

We know that the angle of the minimum fringe in the single-slit experiment is given by

$sin{胃}_m=\frac{m位}{a}$

And in the case of the first minimum fringe, m =1;

$sin{胃}_1=\frac{位}{a}$

Solving for a,

$a=\frac{位}{sin{胃}_1}$

Substitute the given,

$$\begin{gathered} a=\frac{580}{sin{90}^{鈭�}} \\ a=580nm \end{gathered}$$

We know that the intensity is given by

$I=I_0{\left(\frac{\sin {\displaystyle }\frac{尾}{2}}{\frac{尾}{2}}\right)}^2$

We also know that

$尾=\frac{2蟺a\sin 胃}{位}$

Substitute into the intensity formula above, noting that 2 cancels;

$I=I_0{\left(\frac{\sin {\displaystyle }\left(\frac{蟺a\sin 胃}{位}\right)}{\frac{蟺a\sin 胃}{位}}\right)}^2$

To find the ratio of the intensity at$胃=45.0^{鈭�}$ to the ratio of the intensity at $胃=0^{鈭�}$, we need to find$\frac{I}{I_0}$ since the intensity at an angle is the maximum (To). Hence,

$\frac{I}{I_0}={\left(\frac{\sin {\displaystyle }\left(\frac{蟺a\sin 胃}{位}\right)}{\frac{蟺a\sin 胃}{位}}\right)}^2$

Noting that a =$位$as we found in Step 3 above. So,

$\frac{I}{I_0}={\left(\frac{\sin {\displaystyle }(蟺\sin 胃)}{蟺a\sin 胃}\right)}^2$

Substitute the given and note that$45.0掳=\frac{\mathrm{蟺}}{4}$

Remember to convert your calculator to RAD mode.

$$\begin{gathered} \frac{I}{I_0}={\left(\frac{\sin \left(蟺\sin \frac{蟺}{4}\right)}{蟺a\sin \frac{蟺}{4}}\right)}^2 \\ \frac{I}{I_0}=0.128 \end{gathered}$$