91影视

a) When a simple right triangle is constructed, the hypotenuse is the ray leading to the limit of the central maximum, that is, the beginning of the first dark circle, the catheter in front of the small angle is clearly visible. It's also important to remember that when we refer to this little angle, we're referring to its value.

$\tan \left({胃}_1\right)=\frac{R}{L}=\frac{桅}{2L}$

As a consequence, we can compute our circumference in relation of both the angle as follows:

$\mathrm{桅}=2L\tan 鈦�\left({胃}_1\right)$

Keep in mind that we can get the angle as from darker circles criteria as

${胃}_1=\frac{1.22位}{D}$

$$\begin{gathered} \mathrm{桅}=2L\tan 鈦�\left(\frac{1.22位}{D}\right) \\ =2L\tan 鈦�\left(\frac{1.22c}{DV}\right) \end{gathered}$$ ($位=\frac{c}{谓}$)

$$\begin{gathered} \mathrm{桅}=2脳3脳{10}^4\tan 鈦�\left(\frac{1.22脳3脳{10}^8}{2脳1.2脳{10}^{10}\mathrm{V}}\right) \\ =915\mathrm{m} \end{gathered}$$

(Numerically)

b) We can determine the rate based on the power and diameter

$I=\frac{P}{A}=\frac{P}{蟺R^2}=\frac{4P}{蟺{\mathrm{桅}}^2}$

localid="1649147611891" $I=\frac{4脳1脳{10}^5}{蟺脳{915}^2}=0.152\mathrm{W}/{\mathrm{m}}^2$(Numerically)