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Answer the following question in detail. Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.

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Question

Answer the following question in detail.

Derive an expression for the magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.

Answer in Brief
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Solution

  1. Figure (a) shows the visual angle α made by an object when kept at the least distance of distinct vision (D = 25 cm). Without an optical instrument, this is the greatest possible visual angle as we cannot get the object closer than this.

              (a) Visual Angle α


                   (b) Visual Angle β
  2. Figure (b) shows a convex lens forming an erect, virtual, and magnified image of the same object when placed within the focus.
  3. The visual angle β of the object and the image, in this case, is the same. However, this time the viewer is looking at the image which is not closer than D. Hence the same object is now at a distance smaller than D.
  4. Angular magnification or magnifying power, in this case, is given by
    M = `"Visual angle of the image"/"Visual angle of the object at D"=β/α`
    For small angles
    M = `β/α≈(tan(β))/(tan(α))=("BA/PA")/"BA/D"="D"/"u"`
  5. For maximum magnifying power, the image should be at D. For a thin lens, considering thin lens formula, `1/"f"=1/"v"-1/"u"`
    In case of simple microscope,
    f = +f, v = – D, u = – u and M = Mmax
    ∴ `1/"f"=1/-"D"-1/-"u"`
    ∴ `"D"/"f"="D"/-"D"+"D"/"u"`
    As, M = `"D"/"u"`,
    Mmax = `1+"D"/"f"`
  6. For minimum magnifying power, v = ∞ and u = f (numerically)
    ∴ Mmin = `"D"/"u"="D"/"f"`
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Chapter 9: Optics - Exercises [Page 186]

APPEARS IN

Balbharati Physics [English] Standard 11 Maharashtra State Board
Chapter 9 Optics
Exercises | Q 3. (viii) | Page 186

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