Question

A reflection telescope is build with a $20cm$diameter mirror having a $1.00m$focal length. it is used with a $10X$eyepiece. what are

(a) the magnification and

(b) the $f-number$of the telescope?

Short answer

(a) $42km/s,$

(b)$1800km/s.$

Step-by-step solution

part (a) Step 1: Given information

reflection telescope = $20cm$

$focallength=1.00m$

using with a $10x$ eyepiece we need to find magnification

part (a) Step 2: simplify

Since the energy is conserved, all the energy due to the potential drop will be converted to kinetic energy

This means that

$q∆V=\frac{m{v}^2}{2}$

from this, we can find the speed of the particles to be

$\nu =\sqrt{\frac{2q∆V}{m}}$

The absolute value of the charges will be the same, $e$. Also, the product $e∆V$will not change the sign since to accelerate two opposite charges we will have to apply opposite polarity accelerating potential differences. having said that we will numerically have

for the case of the portion

$\nu =\sqrt{\frac{2.1.6.{10}^{-19}.9}{.1.67.{10}^{-27}}}=4.2.{10}^{4}m/s$

part (b) Step 3: Given information

reflection telescope = $20cm$

$focallength=1.00m$

using with a$10X$eyepiece we need to find $f-number$of the telescope

Calculation

For the case of the electrons

$\nu =\sqrt{\frac{2.1.6.{10}^{-19}.9}{9.1.{10}^{-31}}}=1.8.{10}^{6}m/s$