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If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, then the focal length of the eye-piece should be close to
- a)22 mm
- b)12 mm
- c)33 mm
- d)2 mm
Correct answer is option 'A'. Can you explain this answer?
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If we need a magnification of 375 from a compound microscope of tube l...
Given:
- Magnification required (M) = 375
- Length of the compound microscope tube (L) = 150 mm
- Focal length of the objective lens (fo) = 5 mm
To find:
Focal length of the eyepiece lens (fe)
Solution:
The total magnification (M) of a compound microscope is given by the product of the magnification of the objective lens (Mo) and the magnification of the eyepiece lens (Me).
Total Magnification (M) = Mo × Me
Step 1: Calculate the magnification of the objective lens.
Given that the magnification of the objective lens (Mo) can be calculated using the formula:
Mo = 1 + (D / fo)
Where D is the distance of the object from the objective lens and fo is the focal length of the objective lens.
In this case, D is the length of the compound microscope tube (L) = 150 mm.
Mo = 1 + (150 mm / 5 mm)
Mo = 1 + 30
Mo = 31
Step 2: Calculate the magnification of the eyepiece lens.
Given that the total magnification (M) is the product of the magnification of the objective lens (Mo) and the magnification of the eyepiece lens (Me).
M = Mo × Me
In this case, M = 375 and Mo = 31.
Substituting the values in the equation:
375 = 31 × Me
Me = 375 / 31
Me ≈ 12.097
Step 3: Calculate the focal length of the eyepiece lens.
Given that the magnification of the eyepiece lens (Me) can be calculated using the formula:
Me = 1 + (D / fe)
In this case, D is the length of the compound microscope tube (L) = 150 mm.
12.097 = 1 + (150 mm / fe)
fe = 150 mm / 11.097
fe ≈ 13.5 mm
Conclusion:
The focal length of the eyepiece lens (fe) should be close to 13.5 mm, which is closest to option B: 12 mm. However, it is important to note that the given answer of option A: 22 mm is not correct.
- Magnification required (M) = 375
- Length of the compound microscope tube (L) = 150 mm
- Focal length of the objective lens (fo) = 5 mm
To find:
Focal length of the eyepiece lens (fe)
Solution:
The total magnification (M) of a compound microscope is given by the product of the magnification of the objective lens (Mo) and the magnification of the eyepiece lens (Me).
Total Magnification (M) = Mo × Me
Step 1: Calculate the magnification of the objective lens.
Given that the magnification of the objective lens (Mo) can be calculated using the formula:
Mo = 1 + (D / fo)
Where D is the distance of the object from the objective lens and fo is the focal length of the objective lens.
In this case, D is the length of the compound microscope tube (L) = 150 mm.
Mo = 1 + (150 mm / 5 mm)
Mo = 1 + 30
Mo = 31
Step 2: Calculate the magnification of the eyepiece lens.
Given that the total magnification (M) is the product of the magnification of the objective lens (Mo) and the magnification of the eyepiece lens (Me).
M = Mo × Me
In this case, M = 375 and Mo = 31.
Substituting the values in the equation:
375 = 31 × Me
Me = 375 / 31
Me ≈ 12.097
Step 3: Calculate the focal length of the eyepiece lens.
Given that the magnification of the eyepiece lens (Me) can be calculated using the formula:
Me = 1 + (D / fe)
In this case, D is the length of the compound microscope tube (L) = 150 mm.
12.097 = 1 + (150 mm / fe)
fe = 150 mm / 11.097
fe ≈ 13.5 mm
Conclusion:
The focal length of the eyepiece lens (fe) should be close to 13.5 mm, which is closest to option B: 12 mm. However, it is important to note that the given answer of option A: 22 mm is not correct.
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Community Answer
If we need a magnification of 375 from a compound microscope of tube l...
The magnification is given by
⇒ fe = 22 mm
⇒ fe = 22 mm
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If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, then the focal length of the eye-piece should be close toa)22 mmb)12 mmc)33 mmd)2 mmCorrect answer is option 'A'. Can you explain this answer?
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