
A Galileo telescope has an objective of focal length \[100\,{\text{cm}}\] and magnifying power \[50\]. The distance between the two lenses in normal adjustment will be?
A. \[150\,{\text{cm}}\]
B. \[100\,{\text{cm}}\]
C. \[98\,{\text{cm}}\]
D. \[200\,{\text{cm}}\]
Answer
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Hint: Use the formula for the magnifying power of a telescope. This formula gives the relation between the magnifying power of a telescope, focal length of the objective lens and focal length of the eyepiece lens. Substitute all the given values in this formula and calculate the focal length of the eyepiece lens. Then the distance between the two lenses in normal adjustment is the sum of the focal lengths of the objective lens and eyepiece lens which will give the final answer.
Formula used:
The magnifying power \[m\] of a telescope is given by
\[m = - \dfrac{{{f_o}}}{{{f_e}}}\] …… (1)
Here, \[{f_o}\] is focal length of the objective lens and ef is focal length of the eyepiece lens.
Complete step by step answer:
We have given that focal length of the object lens is \[100\,{\text{cm}}\].
\[{f_o} = 100\,{\text{cm}}\]
The magnifying power of the telescope is 50.
\[m = 50\]
Let us first determine the focal length of the eyepiece lens.
We can calculate the focal length of the eyepiece lens using equation (1).
Rearrange equation (1) for focal length pf the eyepiece lens.
\[{f_e} = - \dfrac{{{f_o}}}{m}\]
Substitute \[100\,{\text{cm}}\] for \[{f_o}\] and \[50\] for \[m\] in the above equation.
\[{f_e} = - \dfrac{{100\,{\text{cm}}}}{{50}}\]
\[ \Rightarrow {f_e} = - 2\,{\text{cm}}\]
Hence, the focal length of the eyepiece lens is \[ - 2\,{\text{cm}}\].
The distance \[d\] between the two lenses that is an objective lens and eyepiece lens for a telescope in normal adjustment is the sum of focal length of the object lens and eyepiece lens.
\[d = {f_o} + {f_e}\]
Substitute \[100\,{\text{cm}}\] for \[{f_o}\] and \[ - 2\,{\text{cm}}\] for \[{f_e}\] in the above
equation.
\[d = \left( {100\,{\text{cm}}} \right) + \left( { - 2\,{\text{cm}}} \right)\]
\[ \Rightarrow d = 98\,{\text{cm}}\]
Therefore, the distance between the two lenses in normal adjustment is \[98\,{\text{cm}}\].
Hence, the correct option is C.
Note: The students should keep in mind that the negative sign of the focal length of the eyepiece lens is used because the eyepiece lens used in the telescope is concave and focal length of the concave lens is negative. The objective lens used in the telescope is a convex lens and focal length of the convex lens is positive.
Formula used:
The magnifying power \[m\] of a telescope is given by
\[m = - \dfrac{{{f_o}}}{{{f_e}}}\] …… (1)
Here, \[{f_o}\] is focal length of the objective lens and ef is focal length of the eyepiece lens.
Complete step by step answer:
We have given that focal length of the object lens is \[100\,{\text{cm}}\].
\[{f_o} = 100\,{\text{cm}}\]
The magnifying power of the telescope is 50.
\[m = 50\]
Let us first determine the focal length of the eyepiece lens.
We can calculate the focal length of the eyepiece lens using equation (1).
Rearrange equation (1) for focal length pf the eyepiece lens.
\[{f_e} = - \dfrac{{{f_o}}}{m}\]
Substitute \[100\,{\text{cm}}\] for \[{f_o}\] and \[50\] for \[m\] in the above equation.
\[{f_e} = - \dfrac{{100\,{\text{cm}}}}{{50}}\]
\[ \Rightarrow {f_e} = - 2\,{\text{cm}}\]
Hence, the focal length of the eyepiece lens is \[ - 2\,{\text{cm}}\].
The distance \[d\] between the two lenses that is an objective lens and eyepiece lens for a telescope in normal adjustment is the sum of focal length of the object lens and eyepiece lens.
\[d = {f_o} + {f_e}\]
Substitute \[100\,{\text{cm}}\] for \[{f_o}\] and \[ - 2\,{\text{cm}}\] for \[{f_e}\] in the above
equation.
\[d = \left( {100\,{\text{cm}}} \right) + \left( { - 2\,{\text{cm}}} \right)\]
\[ \Rightarrow d = 98\,{\text{cm}}\]
Therefore, the distance between the two lenses in normal adjustment is \[98\,{\text{cm}}\].
Hence, the correct option is C.
Note: The students should keep in mind that the negative sign of the focal length of the eyepiece lens is used because the eyepiece lens used in the telescope is concave and focal length of the concave lens is negative. The objective lens used in the telescope is a convex lens and focal length of the convex lens is positive.
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