Question

Isaac Newton, having convinced himself (erroneously as it turned out) that chromatic aberration is an inherent property of refracting telescopes, invented the reflecting telescope, shown schematically in Fig. 34-59. He presented his second model of this telescope, with a magnifying power of 38, to the Royal Society (of London), which still has it. In Fig. 34-59, incident light falls, closely parallel to the telescope axis, on the objective mirror. After reflection from the small mirror (the figure is not to scale), the rays form a real, inverted image in the focal plane (the plane perpendicular to the line of sight, at focal point F). This image is then viewed through an eyepiece. (a) Show that the angular magnification for the device is given by Eq. 34-15:

${\mathit{m}}_{\mathbf{\theta }}\mathbf{=}{\mathit{f}}_{\mathbf{o}\mathbf{b}}\mathbf{/}{\mathit{f}}_{\mathbf{e}\mathbf{y}}$

${\mathit{f}}_{\mathbf{o}\mathbf{b}}$

the focal length of the objective is a mirror and

${\mathit{f}}_{\mathbf{ey}}$is that of the eyepiece.

(b) The 200 in. mirror in the reflecting telescope at Mt. Palomar in California has a focal length of 16.8 m. Estimate the size of the image formed by this mirror when the object is a meter stick 2.0 km away. Assume parallel incident rays. (c) The mirror of a different reflecting astronomical telescope has an effective radius of curvature of 10 m (“effective” because such mirrors are ground to a parabolic rather than a spherical shape, to eliminate spherical aberration defects). To give an angular magnification of 200, what must be the focal length of the eyepiece?

Short answer

1. angular magnification$m_{\theta }$ is equal to$\frac{f_{ob}}{f_{ey}}$
2. Size of the image is$8.4m$
3. Focal length of eyepiece is$2.5cm$

Step-by-step solution

Given information:

$r=10\mathrm{m}$

Focal length of the mirror is $16.8m$

$m_{\theta }=200$

Understanding the given information

The problem is based on the principle of refracting telescopes. It is a type of optical telescope that uses a lens as its objective to form an image. It also deals with the angular magnification of the telescope. It is the ratio of the tangents of the angles subtended by an object and its image when measured from a given point in the instrument, as with magnifiers and binoculars.

Formula: ${\mathit{m}}_{\mathbf{\theta }}\mathbf{=}{\mathit{f}}_{\mathbf{ob}}\mathbf{/}{\mathit{f}}_{\mathbf{ey}}$ (i)

Where ${\mathit{v}}_{\mathbf{0}}$and ${\mathit{v}}_f$ are the initial and final velocities.

Explanation

(a)

If we swap out the objective lens in Fig. 34-21 for an objective mirror, the concept of the refracting telescope from the textbook applies to the Newtonian configuration (with the light incident on it from the right). This may imply that the head in Fig. 34-21 would obstruct the incident light, which is why Newton included the mirror M' in his design (to move the head and eyepiece out of the way of the incoming light). The advantage of categorizing mirrors and lenses according to their focal lengths is that, in situations like these, it is simple to transfer the findings of the objective-lens telescope to the objective-mirror telescope by simply swapping out one positive f device for another positive f device.

As a result, a concave mirror must be used in place of the converging lens that serves as the objective of Fig. 34-21 (much as Newton did in Fig. 34-58).

The refracting telescope, which produces and angular magnification $m_{\theta }$given by,

$m_{\theta }=-\frac{f_{ob}}{f_{ey}}$ (ii)

Equation (ii) applies equally as well to the Newtonian telescope:$m_{\theta }=f_{ob}/f_{ey}$

(b) To estimate the size of the image formed by this mirror when the object is a meter stick 2.0 km away 

A meter stick at a distance of 2000 m subtends an angle of

$\begin{array}{c}{\theta }_{stick}=\frac{1\mathrm{m}}{2000\mathrm{m}}\\ =0.0005\mathrm{rad}\end{array}$

Thus, the size of the image formed by the mirror is calculated by multiplying this by the mirror focal length gives

$\left(16.8\mathrm{m}\right)\left(0.0005\right)=8.4\mathrm{m}$.

(c) To calculate the focal length of the eyepiece 

The focal length is given by,

$f=\frac{1}{2}r$, where r is the radius of curvature.

With this, we get

$\begin{array}{c}{f}_{ob}=\frac{10}{2}\\ =5.0\mathrm{m}\end{array}$

Applying this in equation (i),

$m_{\theta }=\frac{f_{ob}}{f_{ey}}$

$\begin{array}{c}{f}_{ey}=\frac{f_{ob}}{m_{\theta }}\\ =\frac{5}{200}\\ =2.5 \mathrm{cm}\end{array}$

Thus, the focal length of eyepiece is 2.5 cm.