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Geometrical Optics question

2020 · 7 Jan · Shift 1 · Q52
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  5. /2020 · 7 Jan · Shift 1 · Q52

Geometrical Optics question

2020 · 7 Jan · Shift 1 · Q52

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :
  1. A
    22 mm
  2. B
    12 mm
  3. C
    33 mm
  4. D
    2 mm
View written solutionFree

Correct answer: A

  1. Magnifying power of a compound microscope

For a compound microscope in normal adjustment (final image at infinity), the magnifying power is

M=(Lfo)(Dfe)M = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)M=(fo​L​)(fe​D​)

where:

  • LLL = tube length
  • fof_ofo​ = focal length of objective
  • fef_efe​ = focal length of eyepiece
  • D=25 cm=250 mmD = 25\ \text{cm} = 250\ \text{mm}D=25 cm=250 mm

Given:

M=375,L=150 mm,fo=5 mmM = 375, \quad L = 150\ \text{mm}, \quad f_o = 5\ \text{mm}M=375,L=150 mm,fo​=5 mm

We need to find fef_efe​.

  1. Substitute the values

375=(1505)(250fe)375 = \left(\frac{150}{5}\right)\left(\frac{250}{f_e}\right)375=(5150​)(fe​250​)

First,

1505=30\frac{150}{5} = 305150​=30

So,

375=30⋅250fe375 = 30\cdot \frac{250}{f_e}375=30⋅fe​250​

375=7500fe375 = \frac{7500}{f_e}375=fe​7500​

  1. Solve for fef_efe​

fe=7500375=20 mmf_e = \frac{7500}{375} = 20\ \text{mm}fe​=3757500​=20 mm

  1. Choose the closest option

Calculated focal length:

fe≈20 mmf_e \approx 20\ \text{mm}fe​≈20 mm

Options are:

  • A: 22 mm22\ \text{mm}22 mm
  • B: 12 mm12\ \text{mm}12 mm
  • C: 33 mm33\ \text{mm}33 mm
  • D: 2 mm2\ \text{mm}2 mm

The closest value is 22 mm22\ \text{mm}22 mm.

  1. Final answer

The focal length of the eyepiece should be close to 22 mm22\ \text{mm}22 mm.

So, Option A is correct.

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