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The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece 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
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The magnifying power of a telescope is 9. When it is adjusted for para...
∴ f0 = 9feAlso f0 + fe = 20 (Qfinal image is at infinity)
9 fe + fe = 20, fe = 2 cm, ∴ f0 = 18 cm
Most Upvoted Answer
The magnifying power of a telescope is 9. When it is adjusted for para...
Given:
- Magnifying power of the telescope, M = 9
- Distance between the objective and the eyepiece, d = 20 cm
To find:
- Focal length of the lenses
Formula:
The magnifying power of a telescope is given by the formula:
M = -fo/fe
Where,
- M is the magnifying power
- fo is the focal length of the objective lens
- fe is the focal length of the eyepiece lens
Calculation:
Given that the magnifying power of the telescope is 9, we can write the equation as:
9 = -fo/fe ...(1)
We are also given that the distance between the objective and the eyepiece is 20 cm. In a telescope adjusted for parallel rays, the distance between the objective and the eyepiece is equal to the sum of the focal lengths of the lenses. Therefore, we can write the equation as:
d = fo + fe ...(2)
Solving equations (1) and (2) simultaneously:
Substituting the value of fe from equation (2) into equation (1), we get:
9 = -fo/(d - fo)
Cross multiplying, we get:
9(d - fo) = -fo
Expanding, we get:
9d - 9fo = -fo
Rearranging, we get:
8fo = 9d
Simplifying, we get:
fo = (9/8)d
Given that d = 20 cm, substituting the value, we get:
fo = (9/8)(20) = 22.5 cm
Therefore, the focal length of the objective lens is 22.5 cm. But the options given do not include this value.
Conclusion:
None of the given options are correct.
- Magnifying power of the telescope, M = 9
- Distance between the objective and the eyepiece, d = 20 cm
To find:
- Focal length of the lenses
Formula:
The magnifying power of a telescope is given by the formula:
M = -fo/fe
Where,
- M is the magnifying power
- fo is the focal length of the objective lens
- fe is the focal length of the eyepiece lens
Calculation:
Given that the magnifying power of the telescope is 9, we can write the equation as:
9 = -fo/fe ...(1)
We are also given that the distance between the objective and the eyepiece is 20 cm. In a telescope adjusted for parallel rays, the distance between the objective and the eyepiece is equal to the sum of the focal lengths of the lenses. Therefore, we can write the equation as:
d = fo + fe ...(2)
Solving equations (1) and (2) simultaneously:
Substituting the value of fe from equation (2) into equation (1), we get:
9 = -fo/(d - fo)
Cross multiplying, we get:
9(d - fo) = -fo
Expanding, we get:
9d - 9fo = -fo
Rearranging, we get:
8fo = 9d
Simplifying, we get:
fo = (9/8)d
Given that d = 20 cm, substituting the value, we get:
fo = (9/8)(20) = 22.5 cm
Therefore, the focal length of the objective lens is 22.5 cm. But the options given do not include this value.
Conclusion:
None of the given options are correct.
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The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer?
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The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer?.
The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses area)18 cm, 2 cmb)11 cm, 9 cmc)10 cm, 10 cmd)15 cm, 5 cmCorrect answer is option 'A'. Can you explain this answer?.
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