
The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be 20 cm. The focal length of lenses are
(A) 18 cm, 2 cm
(B) 11 cm, 9 cm
(C) 10 cm, 10 cm
(D) 15 cm, 5 cm
Answer
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Hint
The magnification of a telescope is directly related to the focal length of the objective lens and inversely related to the focal length of the eye piece. The sum of these focal lengths is the distance between the lenses when adjusted for parallel rays.
Formula used: , where is the magnifying power, is the objective lens focal length, is the eyepiece focal length.
, where is the length between the objective piece and the eyepiece.
Complete step by step answer
We start with using the formula for magnification or magnifying power of a telescope, and it is given as
, is the objective lens focal length, is the eyepiece focal length.
According to the question, is 9.
Thus,
Cross-multiplying, we have
The distance between the objective lens and the eyepiece is given as 20 cm.
When a telescope is adjusted for parallel rays, the objective and eyepiece are configured in such a way that the image focal point (focal point at the side of the lens where images are formed) of the objective lens coincides with the object focal point (focal point at the side of the lens closer to the objective lens) of the eye piece.
Therefore, the distance between the lenses is
Substituting into gives
Since ,
Dividing both sides by 10 we get,
Since
Thus,
and
Hence, the correct option is A.
Note
Alternately, from the equation we can conclude our answer from the set of options given by testing the options against this equation as done below
For option A (18 cm, 2 cm):
which corresponds to the equation . Before concluding, let’s ensure it’s the only one.
For option B (11 cm, 9 cm):
Thus, it doesn’t correspond.
For option C (10 cm, 10cm):
, which again doesn’t correspond.
Finally, for option D (15 cm, 5 cm):
also defies the equation
Therefore, we can conclude that is the only equation that obeys.
Hence, option A is the correct answer.
The magnification of a telescope is directly related to the focal length of the objective lens and inversely related to the focal length of the eye piece. The sum of these focal lengths is the distance between the lenses when adjusted for parallel rays.
Formula used:
Complete step by step answer
We start with using the formula for magnification or magnifying power of a telescope, and it is given as
According to the question,
Thus,
Cross-multiplying, we have
The distance between the objective lens and the eyepiece
When a telescope is adjusted for parallel rays, the objective and eyepiece are configured in such a way that the image focal point (focal point at the side of the lens where images are formed) of the objective lens coincides with the object focal point (focal point at the side of the lens closer to the objective lens) of the eye piece.
Therefore, the distance between the lenses
Substituting
Since
Dividing both sides by 10 we get,
Since
Thus,
Hence, the correct option is A.
Note
Alternately, from the equation
For option A (18 cm, 2 cm):
For option B (11 cm, 9 cm):
Thus, it doesn’t correspond.
For option C (10 cm, 10cm):
Finally, for option D (15 cm, 5 cm):
Therefore, we can conclude that
Hence, option A is the correct answer.
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