Chapter 6, Problem 23Q

(a)

To determine

The magnification from a Newtonian reflector with an objective mirror of $20\text{cm}$ $\left(8\text{in}\text{.}\right)$ diameter, $2\text{m}$ focal length, and an eyepiece of focal length $9\text{mm}$.

Answer to Problem 23Q

Solution:

222, usually written as $222\times$.

Explanation of Solution

Given data:

Diameter of the objective mirror is $20\text{cm}$, focal length of the objective mirror is $2\text{m}$, and the focal length of the eyepiece is $9\text{mm}$.

Formula used:

State the expression for the magnification of a reflecting telescope.

$m=\frac{f}{f_o}$

Here, $m$ is the magnification produced by the telescope, $f$ is the focal length of the objective mirror, and $f_o$ is the focal length of the eyepiece.

Explanation:

The focal length of the objective mirror is $2\text{m}$ and the focal length of the eyepiece is $9\text{mm}$.

Since $1\text{m}$ is equal to $1000\text{mm}$, $2\text{m}$ is equal to $2000\text{mm}$.

Refer to the expression for the magnification of a reflecting telescope.

$m=\frac{f}{f_o}$

Substitute $2000\text{mm}$ for f and $9\text{mm}$ for $f_o$.

$\begin{array}{c}m=\frac{2000\text{mm}}{9\text{mm}}\\ =222.22\\ \approx 222\end{array}$

Conclusion:

Hence, the magnification from the reflecting mirror with an objective mirror of focal length $2000\text{mm}$ and an eyepiece of focal length $9\text{mm}$ is $222\times$.

(b)

To determine

The magnification from a Newtonian reflector with an objective mirror of 20 cm $\left(8\text{in}\text{.}\right)$ diameter, $2\text{m}$ focal length, and an eyepiece of focal length $20\text{mm}$.

Answer to Problem 23Q

Solution:

100, usually written as $100\times$.

Explanation of Solution

Given data:

Diameter of the objective mirror is $20\text{ cm}$, focal length of the objective mirror is $2\text{mm}$, and focal length of the eyepiece is $20\text{mm}$.

Formula used:

State the expression for the magnification of a reflecting telescope.

$m=\frac{f}{f_o}$

Here, $m$ is the magnification, $f$ is the focal length of the objective mirror, and $f_o$ is the focal length of the eyepiece.

Explanation:

The diameter of the objective mirror is $20\text{cm}$, the focal length of the objective mirror is $2\text{m}$, and the focal length of the eyepiece is $20\text{mm}$.

Since $1\text{m}$ is equal to $1000\text{mm}$, $2\text{m}$ is equal to $2000\text{mm}$.

Refer to the expression for the magnification of a reflecting telescope.

$m=\frac{f}{f_o}$

Substitute $2000\text{mm}$ for $f$ and $20\text{mm}$ for $f_o$.

$\begin{array}{c}m=\frac{2000\text{mm}}{20\text{mm}}\\ =100\end{array}$

Conclusion:

Hence, the magnification from a reflecting mirror with an objective mirror of focal length $2000\text{mm}$ and an eyepiece of focal length $20\text{mm}$ is $100\times$.

(c)

To determine

The magnification from a Newtonian reflector with an objective mirror of $20\text{cm}$ $\left(8\text{in}\text{.}\right)$ diameter, $2m$ focal length, and an eyepiece of focal length $55\text{mm}$.

Answer to Problem 23Q

Solution:

$36$, usually written as $36\times$.

Explanation of Solution

Given data:

Diameter of the objective mirror is $20\text{ cm}$, focal length of the objective mirror is $2\text{m}$, and focal length of the eyepiece is $55\text{mm}$.

Formula used:

State the expression for the magnification of a reflecting telescope.

$m=\frac{f}{f_o}$

Here, $m$ is the magnification, $f$ is the focal length of the objective mirror, and $f_o$ is the focal length of the eyepiece

Explanation:

The diameter of the objective mirror is $20\text{cm}$, focal length of the objective mirror is $2\text{m}$, and focal length of the eyepiece is $55\text{mm}$.

Since $1\text{m}$ is equal to $1000\text{mm}$, $2\text{m}$ is equal to $2000\text{mm}$.

Refer to the expression for the magnification of a reflecting telescope.

$m=\frac{f}{f_o}$

Substitute $2000\text{mm}$ for $f$ and $55\text{mm}$ for $f_o$.

$\begin{array}{c}m=\frac{2000\text{mm}}{55\text{mm}}\\ =36.36\\ \approx 36\end{array}$

Conclusion:

Hence, the magnification from a reflecting mirror with an objective mirror of focal length $2000\text{mm}$ and an eyepiece of focal length $55\text{mm}$ is $36\times$.

(d)

To determine

The telescope’s diffraction-limited angular resolution when orange light of wavelength 600 nm is used.

Answer to Problem 23Q

Solution:

$0.075\text{arcsec}$

Explanation of Solution

Given data:

Wavelength of the light used is $600\text{nm}$ and the diameter of the objective mirror of the telescope is $2\text{m}$.

Formula used:

State the expression for the diffraction-limited angular diffraction of a telescope.

$\theta =2.5\times {10}^5\left(\frac{\lambda }{D}\right)$

Here, $\theta$ is the diffraction-limited angular diffraction of a telescope in arcsecond, $\lambda$ is the wavelength of light in meters, and $D$ is the diameter of the telescope’s objective in meters.

Explanation:

The diameter of the objective mirror of the telescope is $20\text{cm}$ and the wavelength of the light used is $600\text{nm}$.

Since $1\text{nm}$ is equal to $1\times {10}^{-9}$ meter, $600\text{nm}$ is equal to $6\times {10}^{-7}\text{m}$.

Refer to the expression for the diffraction-limited angular diffraction of a telescope.

$\theta =2.5\times {10}^5\left(\frac{\lambda }{D}\right)$

Substitute $6\times {10}^{-7}\text{m}$ for $\lambda$ and $2\text{m}$ for $D$.

$\begin{array}{c}\theta =2.5\times {10}^5\left(\frac{6\times {10}^{-7}\text{m}}{2\text{m}}\right)\\ =0.075\text{arcsec}\end{array}$

Conclusion:

Hence, the diffraction-limited angular resolution of a telescope when a light of wavelength $600\text{nm}$ is used is $0.075$ arcsecond.

(e)

To determine

Whether it’s possible to achieve diffraction-limited angular resolution when orange light with a wavelength of $600\text{nm}$ is used, if the telescope is taken to the summit of Mauna Kea.

Answer to Problem 23Q

Solution:

No, we can’t achieve angular resolution, which is equal to $0.075$ arcsecond, because after the resolution reaches a certain limit, the images obtained will be blurred because of the atmosphere.

Explanation of Solution

Introduction:

Diffraction-limited angular resolution of a telescope indicates the limit of the telescope up to which it can generate clear images, assuming that diffraction is occurring while the observation is taking place. Any device with a high angular resolution can differentiate minute details of different objects.

Explanation:

As Earth is surrounded by atmosphere, anything that enters through the atmosphere has to pass through the gases present in it.

In the case of diffraction-limited angular resolution of a telescope, a large amount of light gets diffracted before entering the objective mirror of a telescope, therefore, limiting the resolution power of the telescope.

Conclusion:

Hence, we can’t achieve a value of $0.075$ arcsecond for a diffraction-limited angular resolution even if we take our telescope to the summit of Mauna Kea.