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In a microscope the objective is having focal length $f_o = 2 \text{ cm}$ and eye-piece is having focal length $f_e = 4 \text{ cm}$. The tube length is $32 \text{ cm}$. The magnification produced by this microscope for normal adjustment is ____________.

Microscope Magnification Calculation for Normal Adjustment

This solution details the calculation for the angular magnification of a compound microscope when the final image is formed at infinity (normal adjustment).

Given Data

  • Objective focal length, $f_o = 2 \text{ cm}$
  • Eyepiece focal length, $f_e = 4 \text{ cm}$
  • Tube length, $L = 32 \text{ cm}$
  • Near point of distinct vision, $D = 25 \text{ cm}$ (assumed standard value)

Magnification Formula

For a microscope adjusted for normal viewing (final image at infinity), the total angular magnification ($M$) is given by the product of the objective lens magnification ($M_o$) and the eyepiece lens magnification ($M_e$):

$M = M_o \times M_e$

Where:

  • Objective magnification: $M_o = \frac{L}{f_o}$
  • Eyepiece magnification for normal adjustment: $M_e = \frac{D}{f_e}$

Substituting these into the total magnification formula gives:

$M = \frac{L}{f_o} \times \frac{D}{f_e}$

Calculation Steps

  1. Calculate Objective Magnification ($M_o$): $M_o = \frac{L}{f_o} = \frac{32 \text{ cm}}{2 \text{ cm}} = 16$
  2. Calculate Eyepiece Magnification ($M_e$) for Normal Adjustment: $M_e = \frac{D}{f_e} = \frac{25 \text{ cm}}{4 \text{ cm}} = 6.25$
  3. Calculate Total Magnification ($M$): $M = M_o \times M_e = 16 \times 6.25 = 100$

Result

The magnification produced by the microscope for normal adjustment is 100.

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