Question

A small telescope has an objective lens of focal length 150cm and an eye piece of focal length 5cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment? If this telescope is used to view a $100 m$ tall tower $3 km$ away, what is the height of the image of the tower formed by the objective lens?

Answer 1

Hint: We know that the value of distance of distinct vision is $25 cm$. Use the formula for magnification of a telescope and determine the magnification using the given quantities. To determine the height of the image of the tower in the telescope, we can find the angle subtended by the tower as seen from our eyes. This angle is the same for the angle subtended by the image.

Formula used:
The formula for magnification of the telescope,
\[m = - \dfrac{{{f_0}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)\]
Here, D is the distance of distinct vision, \[{f_0}\] focal length of objective lens, and \[{f_e}\] focal length of eyepiece.

Complete step by step answer:
We have given focal length of objective lens\[{f_0} = 150\,cm\], focal length of eyepiece \[{f_e} = 5\,cm\], height of tower \[H = 100\,m\]and distance of tower \[d = 3\,km\].
We have the formula for magnification of the telescope,
\[m = - \dfrac{{{f_0}}}{{{f_e}}}\left( {1 + \dfrac{{{f_e}}}{D}} \right)\]
Here, D is the distance of distinct vision and its value is 25 cm.
We substitute\[{f_0} = 150\,cm\], \[{f_e} = 5\,cm\] and \[D = 25\,cm\] in the above equation.
\[m = - \dfrac{{150}}{5}\left( {1 + \dfrac{5}{{25}}} \right)\]
\[ \Rightarrow m = - \dfrac{{150}}{5}\left( {1.2} \right)\]
\[ \Rightarrow m = - 36\]
Therefore, the magnification of this telescope is \[ - 36\]. The negative sign implies that the image is inverted.
If we observe the tower with our eyes, then we can express the angle subtended by the height of tower as follows,
\[\tan {\theta _1} = \dfrac{H}{d}\]
We substitute 100 m for H and 3000 m for d in the above equation.
\[\tan {\theta _1} = \dfrac{{100}}{{3000}}\]
\[ \Rightarrow \tan {\theta _1} = \dfrac{1}{{30}}\] …… (1)
Now we can express the angle subtended by the tower at the telescope as follows,
\[\tan {\theta _2} = \dfrac{h}{{{f_o}}}\] …… (2)
Here, h is the height of the image of the tower in the telescope.
We know that the angle subtended by the tower \[\tan {\theta _1}\] is the same as the angle subtended by the tower at the telescope. Therefore, we can write,
\[\dfrac{h}{{{f_o}}} = \dfrac{1}{{30}}\]
\[ \Rightarrow h = \dfrac{{{f_o}}}{{30}}\]
We substitute 150 cm for \[{f_o}\] in the above equation.
\[h = \dfrac{{150}}{{30}}\]
\[ \therefore h = 5\,cm\]

Therefore, the height of the image is 5 cm.

Note:To solve this type of question, students should remember the value of distance of distinct vision as it may not be given in the question. In case your sign of height of image is negative, you should know the image of that object is inverted. To determine the height of the object, you can make the geometric construction of object height and object distance. Then using trigonometric formulation, you can determine the angle subtended by the object.