A small telescope has an objective lens of focal length 140cm andan eyepiece of focal length 5.0cm. What is the magnifying power ofthe telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final imageis at infinity)? (b) the final image is formed at the least distance of distinct vision(25cm)?
Given: The focal length of eyepiece is 5.0 cm and the focal length of objective is 140 cm .
a)
When the telescope is in normal adjustment, its magnifying power is given as,
m = f o f e
Where, f e is the focal length of eyepiece and f o is the focal length of objective.
By substituting the given values in the above expression, we get
m = 140 5 = 28
Thus, when the telescope is in normal adjustment, its magnifying power is 28 .
b)
When the final image is formed at least distance for distinct vision, the magnifying power of the telescope is given as,
m = f o f e [ 1 + f e d ]
Where, d is the least distance for distinct vision.
By substituting the given values in the above expression, we get
m = 140 5 [ 1 + 5 25 ] = 28 [ 1 + 0.2 ] = 28 × 1.2 = 33.6
Thus, when the final image is formed at least distance for distinct vision, the magnifying power of the telescope is 33.6 .
Given: The focal length of eyepiece is
a)
When the telescope is in normal adjustment, its magnifying power is given as,
Where,
By substituting the given values in the above expression, we get
Thus, when the telescope is in normal adjustment, its magnifying power is
b)
When the final image is formed at least distance for distinct vision, the magnifying power of the telescope is given as,
Where,
By substituting the given values in the above expression, we get
Thus, when the final image is formed at least distance for distinct vision, the magnifying power of the telescope is
