Chapter 27, Problem 69P

(a)

To determine

The angular resolution of VLA.

Answer to Problem 69P

Angular resolution is $1.50\text{arc}\text{sec}$.

Explanation of Solution

Write the Equation to find the wavelength.

    $\lambda =\frac{c}{f}$

Here, $c$ is the speed of light in vaccum, $f$ is the freuency and $\lambda$ is the wavelength.

Write the equation to find the angular resolution of VLA.

    ${\theta }_{\min }=1.22\frac{\lambda }{D}$        (I)

Here, ${\theta }_{\min }$ is the angular resolution of VLA and $D$ is the diameter of aperture.

Conclusion:

Substitute $3.00\times {10}^8\text{m}/\text{s}$ for $c$ and $1.40\text{GHz}$ for $f$ in the equation for $\lambda$.

    $\begin{array}{c}\lambda =\frac{3.00\times {10}^8\text{m}/\text{s}}{1.40\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)}\\ =0.214\text{m}\end{array}$

Substitute $0.214\text{m}$ for $\lambda$ and $36.0\text{km}$ for $D$ in the equation for ${\theta }_{\min }$.

    $\begin{array}{c}{\theta }_{\min }=1.22\left(\frac{0.214\text{m}}{36.0\text{km}\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)}\right)\\ =7.26\times {10}^{-6}\text{rad}\left(\frac{\mu \text{rad}}{{10}^{-6}\text{rad}}\right)\\ =7.26\mu \text{rad}\end{array}$

Express the calculated value in $\text{arc}\text{sec}$.

    $7.26\mu \text{rad}\left(\frac{180\times 60\times 60\text{s}}{\pi }\right)=1.50\text{arc}\text{sec}$

Therefore, the angular resolution is $1.50\text{arc}\text{sec}$.

(b)

To determine

The seperation between two clouds at the galaxy center.

Answer to Problem 69P

The seperation will be $0.189\text{ly}$.

Explanation of Solution

Write the equation to find the seperation between two clouds at the galaxy center.

  $d={\theta }_{\min }L$

Here, $d$ is the seperation between two clouds at the galaxy center and $L$ is the distance to the center of galaxy.

Conclusion:

Substitute $1.50\text{arc}\text{sec}$ for ${\theta }_{\min }$ and $29,000\text{ly}$ for $L$ in the above equation.

    $\begin{array}{c}d=\left(1.50\text{arc}\text{sec}\left(\frac{\pi }{180\times 60\times 60\text{s}}\right)\right)\left(26,000\text{ly}\right)\\ =0.189\text{ly}\end{array}$

Therefore, the seperation will be $0.189\text{ly}$.

(c)

To determine

The angular resolution of hawk’s eye.

Answer to Problem 69P

The angular resolution is $50.8\mu \text{rad}$.

Explanation of Solution

The given case is not correct for humans. It is assumed that hawk’s eyesight is only limited by Rayleigh’s criterion.

Conclusion:

Substitute $500\text{nm}$ for $\lambda$ and $12.0\text{mm}$ for $D$ in the equation (I).

    $\begin{array}{c}{\theta }_{\min }=1.22\frac{500\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}{12.0\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =50.8\times {10}^{-6}\text{rad}\left(\frac{1\mu \text{rad}}{{10}^{-6}\text{rad}}\right)\\ =50.8\mu \text{rad}\end{array}$

Therefore, the angular resolution is $50.8\mu \text{rad}$.

(d)

To determine

The distance of separation between whiskers of mouse required for hawk to resolve them.

Answer to Problem 69P

The distance of separation must be $1.52\text{mm}$.

Explanation of Solution

Write the equation to find the separation between whiskers of mouse required for hawk to resolve them.

  $d={\theta }_{\min }L$

Here, $d$ is the distance of separation and $L$ is the distance from ground to mouse.

Conclusion:

Substitute $50.8\times {10}^{-6}\text{rad}$ for ${\theta }_{\min }$ and $30.0\text{m}$ for $L$ in the above equation.

    $\begin{array}{c}d=\left(50.8\times {10}^{-6}\text{rad}\right)\left(30.0\text{m}\right)\\ =1.52\times {10}^{-3}\text{m}\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =1.52\text{mm}\end{array}$   

Therefore, the distance of separation must be $1.52\text{mm}$.