Question

(a) What is the angular separation of two stars if their images are barely resolved by the Thaw refracting telescope at the Allegheny Observatory in Pittsburgh? The lens diameter is 76 cm and its focal length is 14 m. Assume $\mathit{\lambda }\mathbf{=}\mathbf{550}\mathbf{}\mathbf{\text{nm}}$. (b) Find the distance between these barely resolved stars if each of them is 10 light-years distant from Earth. (c) For the image of a single star in this telescope, find the diameter of the first dark ring in the diffraction pattern, as measured on a photographic plate placed at the focal plane of the telescope lens. Assume that the structure of the image is associated entirely with diffraction at the lens aperture and not with lens “errors.”

Short answer

1. The angular separation of two stars is $8.8\times {10}^{-7}\text{rad}$.
2. The required distance is $8.4\times {10}^7\text{km}$.
3. The diameter of the first dark ring is$0.025\text{mm}$ .

Step-by-step solution

Concept/Significance of angular separation

The expression to calculate the angular separation is given by,

${\mathit{\theta }}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{\lambda }}{\mathbf{d}}$ …… (1)

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

(a) Find the angular separation of two stars

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

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

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

(b) Find the distance between the barely resolved stars

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.

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

Use the following expression to find D .

$\tan {\theta }_R=\frac{D}{L}$

As ${\theta }_R$is very small, then take $\tan {\theta }_R\approx {\theta }_R$.

$$\begin{gathered} {\theta }_R=\frac{D}{L} \\ D={\theta }_{R}L \end{gathered}$$ ……. (2)

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

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

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

(c) Find the diameter of the first dark ring in the diffraction pattern

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

$d=2{\theta }_{R}L$ …….. (3)

Here, L is the focal length of the lens.

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

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

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