Question

An amateur astronomer wishes to estimate roughly the size of the sun using his crude telescope consisting of an objective lens of focal length 200cm and an eyepiece of focal length 10cm. By adjusting the distance of the eye piece from the objective, he obtains an image of the sun on a screen 40cm behind the eyepiece, the diameter of the sun’s image is measured to be 6.0cm. What is the estimate of the sun’s size, given that the average earth-sun distance is $1.5 \times {10^{11}}m$.

Answer 1

Hint: The distance of the object is calculated from the lens formula. Then, diameter of the sun’s object is calculated from magnification and angle subtended by the objective lens gives the size of the sun’s size.

Complete step by step answer:
An amateur astronomer wants to measure the size of the sun using his telescope whose focal length of objective lens be fo = 200cm and focal length of eyepiece be fe = 10cm. A telescope consists of two lenses placed coaxially and one facing the distant object is called objective and has a large aperture and large focal length. The other lens is called eye piece as the eye is placed close to it. Let the distance of the object from the lens be ue and distance of image from the lens be ve = 40cm.
From the lens formula of the telescope,
$\dfrac{1}{{ve}} - \dfrac{1}{{ue}} = \dfrac{1}{{fe}}$
$\implies \dfrac{1}{{40}} - \dfrac{1}{{ue}} = \dfrac{1}{{\left( { - 10} \right)}}$ $\left[ {focal{\text{ }}length{\text{ }}is{\text{ }}taken{\text{ }}as{\text{ }}negative} \right]$
$\implies \dfrac{1}{{ue}} = \dfrac{1}{{40}} - \dfrac{1}{{10}}$
$\implies \dfrac{1}{{ue}} = \dfrac{{1 - 4}}{{40}}$
$\implies \dfrac{1}{{ue}} = \dfrac{{ - 3}}{{40}}$
$\implies ue = \dfrac{{ - 40}}{3}$
Now, magnification produced due to eyepiece $\left( {me} \right) = $ $\dfrac{{|ve|}}{{|ue|}}$
$me = \dfrac{{|40|}}{{\dfrac{{| - 40|}}{{|3|}}}}$
$me = 3$
Also, $magnification = \dfrac{{{\text{ }}diameter{\text{ }}of{\text{ }}size{\text{ }}of{\text{ }}image{\text{ }}\left( {Di} \right)}}{{diameter{\text{ }}of{\text{ }}size{\text{ }}of{\text{ }}object{\text{ }}\left( {Do} \right)}}$
$Do = \dfrac{{Di}}{{me}}$
Do = 6/3 = 2$Do = \dfrac{6}{3}i.e.,2$
The angle subtended by the objective lens with it be $\theta $ and let d be the size of the sun.
$\theta = \dfrac{d}{{1.5 \times {{10}^{11}}}}\left[ {eqn1} \right]$
and also \[\] \[\theta = \dfrac{{Diameter{\text{ }}of{\text{ }}size{\text{ }}of{\text{ }}the{\text{ }}sun}}{{fo}}\left[ {eqn2} \right]\]
From eqn1 and eqn2,
$\dfrac{d}{{1.5 \times {{10}^{11}}}} = \dfrac{{Do}}{{fo}}$
$\implies \dfrac{d}{{1.5 \times {{10}^{11}}}} = \dfrac{2}{{200}}$
$\implies d = \dfrac{1}{{100}} \times 1.5 \times {10^{11}}$
$\therefore d = 1.5 \times {10^9}$ m.

Note:
The focal length in the given instrument i.e., telescope has negative focal length. The negative sign of object distance indicates that the object is in front of the lens and positive sign of image distance shows that the image is formed behind the lens.