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The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece in cm?
- a)30
- b)10
- c)40
- d)20
Correct answer is option 'B'. Can you explain this answer?
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The magnifying power of a telescope with tube length 60 cm is 5. What ...
Concept:
The magnifying power of a telescope is given by:
M = (-) (angular magnification of objective lens) x (angular magnification of eyepiece).
The angular magnification of the objective lens is given by:
m_o = -f_o/f_e
where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece.
The angular magnification of the eyepiece is given by:
m_e = 1 + f_o/f_e
where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece.
Calculation:
Given, tube length of the telescope, L = 60 cm and magnifying power, M = 5.
Let the focal length of the eyepiece be f_e.
The magnifying power of the telescope is given by:
M = (-) (angular magnification of objective lens) x (angular magnification of eyepiece)
Substituting the values of magnifying power and angular magnification of objective lens, we get:
5 = (-) (f_o/f_e) x (1 + f_o/f_e)
Simplifying, we get:
5 = (-) (f_o + f_e)/(f_e)
Multiplying both sides by f_e, we get:
5f_e = (-) (f_o + f_e)
Rearranging, we get:
6f_e = (-) f_o
Substituting the value of f_o from the equation of the objective lens, we get:
6f_e = (-) (L/f_e)
Multiplying both sides by f_e, we get:
6f_e^2 = (-) L
Substituting the given values, we get:
6f_e^2 = (-) 60
Simplifying, we get:
f_e^2 = 10
Taking the square root of both sides, we get:
f_e = 3.16
Therefore, the focal length of the eyepiece is 3.16 cm.
Answer:
Option B - 10 cm.
The magnifying power of a telescope is given by:
M = (-) (angular magnification of objective lens) x (angular magnification of eyepiece).
The angular magnification of the objective lens is given by:
m_o = -f_o/f_e
where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece.
The angular magnification of the eyepiece is given by:
m_e = 1 + f_o/f_e
where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece.
Calculation:
Given, tube length of the telescope, L = 60 cm and magnifying power, M = 5.
Let the focal length of the eyepiece be f_e.
The magnifying power of the telescope is given by:
M = (-) (angular magnification of objective lens) x (angular magnification of eyepiece)
Substituting the values of magnifying power and angular magnification of objective lens, we get:
5 = (-) (f_o/f_e) x (1 + f_o/f_e)
Simplifying, we get:
5 = (-) (f_o + f_e)/(f_e)
Multiplying both sides by f_e, we get:
5f_e = (-) (f_o + f_e)
Rearranging, we get:
6f_e = (-) f_o
Substituting the value of f_o from the equation of the objective lens, we get:
6f_e = (-) (L/f_e)
Multiplying both sides by f_e, we get:
6f_e^2 = (-) L
Substituting the given values, we get:
6f_e^2 = (-) 60
Simplifying, we get:
f_e^2 = 10
Taking the square root of both sides, we get:
f_e = 3.16
Therefore, the focal length of the eyepiece is 3.16 cm.
Answer:
Option B - 10 cm.
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The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece in cm?a)30b)10c)40d)20Correct answer is option 'B'. Can you explain this answer?
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