Chapter 6, Problem 23Q

To determine

(a)

Magnification of the telescope for the given conditions.

Answer to Problem 23Q

Magnification of the telescope is $222$.

Explanation of Solution

Given data:

Focal length of the objective lens, $f_{obj}=2\text{m}$

Focal length of the eyepiece lens, $f_{eye}=9\text{mm}$

Formula used:

The magnification power of the telescope is given as

$M=\frac{f_{obj}}{f_{eye}}$

Calculation:

The magnification power of the telescope is given as

$M=\frac{f_{obj}}{f_{eye}}$

Plugging the values in the above equation

$\begin{array}{c}M=\frac{2}{0.009}\\ =222\end{array}$

Conclusion:

Magnification of the telescope is $222$.

To determine

(b)

Magnification of the telescope for the given conditions.

Answer to Problem 23Q

Magnification of the telescope is $100$.

Explanation of Solution

Given data:

Focal length of the objective lens, $f_{obj}=2\text{m}$

Focal length of the eyepiece lens, $f_{eye}=20\text{mm}$

Formula used:

The magnification power of the telescope is given as

$M=\frac{f_{obj}}{f_{eye}}$

Calculation:

The magnification power of the telescope is given as

$M=\frac{f_{obj}}{f_{eye}}$

Plugging the values in the above equation

$\begin{array}{c}M=\frac{2}{0.020}\\ =100\end{array}$

Conclusion:

Magnification of the telescope is $100$.

To determine

(c)

Magnification of the telescope for the given conditions.

Answer to Problem 23Q

Magnification of the telescope is $36$.

Explanation of Solution

Given data:

Focal length of the objective lens, $f_{obj}=2\text{m}$

Focal length of the eyepiece lens, $f_{eye}=55\text{mm}$

Formula used:

The magnification power of the telescope is given as

$M=\frac{f_{obj}}{f_{eye}}$

Calculation:

The magnification power of the telescope is given as

$M=\frac{f_{obj}}{f_{eye}}$

Plugging the values in the above equation

$\begin{array}{c}M=\frac{2}{0.055}\\ =36\end{array}$

Conclusion:

Magnification of the telescope is $36$.

To determine

(d)

The diffraction- limited angular resolution of the telescope for the given conditions.

Answer to Problem 23Q

The diffraction- limited angular resolution of the telescope is $0.75\text{arcsecond}$.

Explanation of Solution

Given data:

Diameter of the objective mirror = 20 cm = 0.20 m

Wavelength, $\lambda =600\text{nm}$

Formula used:

The angular resolution of the telescope (in arc-second) is given by the formula

$\theta =2.5\times {10}^5\times \frac{\lambda }{D}$

Here,

$\lambda$ is the wavelength in meters

$D$ is the diameter of the objective lens in meters

Calculation:

The angular resolution of the telescope is given by the formula

$\theta =2.5\times {10}^5\times \frac{\lambda }{D}$

Plugging the values in the above equation

$\begin{array}{c}\theta =2.5\times {10}^5\times \frac{600\times {10}^{-9}}{0.20}\\ =0.75\text{arcsecond}\end{array}$

Conclusion:

Angular resolution of the telescope is $0.75\text{arcsecond}$.

To determine

(e)

Whether or not the angular resolution of the telescope change if one took the telescope to the summit of Mauna Kea.

Answer to Problem 23Q

The angular resolution of the telescope would be same

Explanation of Solution

Given data:

Focal length of the objective lens, $f_{obj}=2\text{m}$

Wavelength, $\lambda =600\text{nm}$

Formula used:

The angular resolution of the telescope (in arc-second) is given by the formula

$\theta =2.5\times {10}^5\times \frac{\lambda }{D}$

Here,

$\lambda$ is the wavelength in meters

$D$ is the diameter of the objective lens in meters

Calculation:

The angular resolution of the telescope is given by the formula

$\theta =2.5\times {10}^5\times \frac{\lambda }{D}$

We can observe from the above formula that the angular resolution does not depend on the positon of the telescope. Therefore, there would be no change in the angular resolution of the telescope.

Conclusion:

Thus, the angular resolution of the telescope would be same even if one took the telescope to the summit of Mauna Kea.