Question
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
- Calculate Objective Magnification ($M_o$): $M_o = \frac{L}{f_o} = \frac{32 \text{ cm}}{2 \text{ cm}} = 16$
- Calculate Eyepiece Magnification ($M_e$) for Normal Adjustment: $M_e = \frac{D}{f_e} = \frac{25 \text{ cm}}{4 \text{ cm}} = 6.25$
- 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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