91影视

$r=10\mathrm{m}$

Focal length of the mirror is $16.8m$

$m_{胃}=200$

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

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

(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 $m_{胃}$given by,

$m_{胃}=-\frac{f_{ob}}{f_{ey}}$ (ii)

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

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

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

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

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

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),

$m_{胃}=\frac{f_{ob}}{f_{ey}}$

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

Thus, the focal length of eyepiece is 2.5 cm.