91影视

The expression to calculate the angular separation is given by,

${\mathit{胃}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{位}}{\mathbf{d}}$ 鈥︹�� (1)

Here, d is the diameter of the aperture, and $\mathit{位}$ is wavelength.

Substitute$76\text{cm}$ for d, and $550\text{nm}$for $位$in equation (1).

$$\begin{gathered} {胃}_R=\frac{1.22\left(550\text{nm}\right)}{\left(76\text{cm}\right)} \\ =\frac{1.22\left(550\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{\left(76\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)} \\ =8.8脳{10}^{-7}\text{rad} \end{gathered}$$

Therefore, the angular separation of two stars is $8.8脳{10}^{-7}\text{rad}$.

Let D be the size of the star that lens resolve, and L be the distance between lens and star.

Convert the distance between lens and star in meter.

$$\begin{gathered} L=\left(10\text{ly}\right)\left(\frac{9.46脳{10}^{15}\text{m}}{1\text{ly}}\right) \\ =94.6脳{10}^{15}\text{m} \end{gathered}$$

Use the following expression to find D .

$\tan {胃}_R=\frac{D}{L}$

As ${胃}_R$is very small, then take $\tan {胃}_R鈮�{胃}_R$.

$$\begin{gathered} {胃}_R=\frac{D}{L} \\ D={胃}_{R}L \end{gathered}$$ 鈥︹��. (2)

Substitute $8.8脳{10}^{-7}\text{rad}$for${胃}_R$ , and $94.6脳{10}^{15}\text{m}$for L in equation (2).

$$\begin{gathered} D=\left(8.8脳{10}^{-7}\text{rad}\right)\left(94.6脳{10}^{15}\text{m}\right) \\ 鈮�8.4脳{10}^{10}\text{m} \\ =8.4脳{10}^7\text{km} \end{gathered}$$

Therefore, the required distance is $8.4脳{10}^7\text{km}$.

The expression to calculate the diameter of the first dark ring is given by,

$d=2{胃}_{R}L$ 鈥︹��.. (3)

Here, L is the focal length of the lens.

Substitute $8.8脳{10}^{-7}\text{rad}$for ${胃}_R$, and 14 m for L in equation (3).

$$\begin{gathered} d=2\left(8.8脳{10}^{-7}\text{rad}\right)\left(14\text{m}\right) \\ =246.4脳{10}^{-7}\text{m} \\ =\left(246.4脳{10}^{-7}\text{m}\right)\left(\frac{{10}^3\text{mm}}{2\text{m}}\right) \\ 鈮�0.025\text{mm} \end{gathered}$$

Therefore, the diameter of the first dark ring is $0.025\text{mm}$.