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The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)?
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The focal length of the eyepiece and the objective of an astronomical ...
Given:
- Focal length of eyepiece (fe) = 4 cm
- Focal length of objective (fo) = 50 cm
- Least distance of distinct vision (D) = 25 cm
To find:
- Length of the telescope for the adjustment in which eye is most strained
Solution:
1. Magnifying power of the telescope (M):
The magnifying power of a telescope is given by the formula:
M = - (fo/fe)
Substituting the given values, we get:
M = - (50/4)
M = -12.5
The negative sign indicates that the image formed by the telescope is inverted.
2. Total magnification (Mtotal):
The total magnification of a telescope is given by the formula:
Mtotal = M * (1 + D/fe)
Substituting the given values, we get:
Mtotal = -12.5 * (1 + 25/4)
Mtotal = -12.5 * (1 + 6.25)
Mtotal = -12.5 * 7.25
Mtotal = -90.625
3. Length of the telescope (L):
The length of the telescope is given by the formula:
L = (fo + fe) * (1 + Mtotal)
Substituting the given values, we get:
L = (50 + 4) * (1 + (-90.625))
L = 54 * (-89.625)
L ≈ -4846.25 cm
Since the length of the telescope cannot be negative, we take the absolute value:
L ≈ 4846.25 cm
Therefore, the length of the telescope for the adjustment in which the eye is most strained is approximately 4846.25 cm.
Conclusion:
The length of the telescope for the adjustment in which the eye is most strained is approximately 4846.25 cm.
- Focal length of eyepiece (fe) = 4 cm
- Focal length of objective (fo) = 50 cm
- Least distance of distinct vision (D) = 25 cm
To find:
- Length of the telescope for the adjustment in which eye is most strained
Solution:
1. Magnifying power of the telescope (M):
The magnifying power of a telescope is given by the formula:
M = - (fo/fe)
Substituting the given values, we get:
M = - (50/4)
M = -12.5
The negative sign indicates that the image formed by the telescope is inverted.
2. Total magnification (Mtotal):
The total magnification of a telescope is given by the formula:
Mtotal = M * (1 + D/fe)
Substituting the given values, we get:
Mtotal = -12.5 * (1 + 25/4)
Mtotal = -12.5 * (1 + 6.25)
Mtotal = -12.5 * 7.25
Mtotal = -90.625
3. Length of the telescope (L):
The length of the telescope is given by the formula:
L = (fo + fe) * (1 + Mtotal)
Substituting the given values, we get:
L = (50 + 4) * (1 + (-90.625))
L = 54 * (-89.625)
L ≈ -4846.25 cm
Since the length of the telescope cannot be negative, we take the absolute value:
L ≈ 4846.25 cm
Therefore, the length of the telescope for the adjustment in which the eye is most strained is approximately 4846.25 cm.
Conclusion:
The length of the telescope for the adjustment in which the eye is most strained is approximately 4846.25 cm.
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The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)?
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The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)?.
The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The focal length of the eyepiece and the objective of an astronomical telescope are 4cm and 50cm . The length of the telescope for the adjustment in which eye is most stained will be about (least distance of distinct vision is 25 cm)?.
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