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An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the focal length f0 of the eyepiece are
  • a)
    f0 = 45 cm and fe = –9 cm
  • b)
    f0 = 50 cm and fe = 10 cm
  • c)
    f0 = 7.2 cm and fe = 5 cm
  • d)
    f0 = 30 cm and fe = 6 cm.
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
An astronomical telescope has an angular magnification of magnitude 5 ...
In an astronomical telescope when the object and final image are at infinity, M and L are given as shown: Angular magnification M = fo/fe
Seperation between lenses, L = fo +fe
fo +fe= 36  or  5 fe +fe= 36
or fe = 6 cm ... (ii)
∴ fo = 5fe  or  fo = 30 cm
Hence fo = 30 cm,  fe = 6 cm
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Most Upvoted Answer
An astronomical telescope has an angular magnification of magnitude 5 ...
The angular magnification (M) of an astronomical telescope is given by the formula:

M = - (f0 / fe)

where f0 is the focal length of the objective and fe is the focal length of the eyepiece.

Given that the angular magnification is 5, we can substitute this value into the formula:

5 = - (f0 / fe)

Rearranging the formula, we get:

f0 = -5fe

We also know that the separation between the objective and the eyepiece (d) is 36 cm. The formula for the total length of the telescope (L) is given by:

L = f0 + fe

Substituting the value of f0 from the previous equation, we get:

L = -5fe + fe
L = -4fe

Given that the final image is formed at infinity, the total length of the telescope is equal to the separation between the objective and the eyepiece:

L = d
-4fe = 36

Solving for fe, we get:

fe = -36 / -4
fe = 9 cm

Substituting this value into the equation for f0, we can solve for f0:

f0 = -5fe
f0 = -5(9)
f0 = -45

Therefore, the focal length of the objective (f0) is 45 cm and the focal length of the eyepiece (fe) is 9 cm.
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An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the focal length f0 of the eyepiece area)f0 = 45 cm and fe = –9 cmb)f0 = 50 cm and fe = 10 cmc)f0 = 7.2 cm and fe = 5 cmd)f0 = 30 cm and fe = 6 cm.Correct answer is option 'D'. Can you explain this answer?
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An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the focal length f0 of the eyepiece area)f0 = 45 cm and fe = –9 cmb)f0 = 50 cm and fe = 10 cmc)f0 = 7.2 cm and fe = 5 cmd)f0 = 30 cm and fe = 6 cm.Correct answer is option 'D'. 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 An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the focal length f0 of the eyepiece area)f0 = 45 cm and fe = –9 cmb)f0 = 50 cm and fe = 10 cmc)f0 = 7.2 cm and fe = 5 cmd)f0 = 30 cm and fe = 6 cm.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the focal length f0 of the eyepiece area)f0 = 45 cm and fe = –9 cmb)f0 = 50 cm and fe = 10 cmc)f0 = 7.2 cm and fe = 5 cmd)f0 = 30 cm and fe = 6 cm.Correct answer is option 'D'. Can you explain this answer?.
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