© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Formula for Magnification
For a compound microscope with the final image formed at the near point (normal adjustment), the magnifying power (M.P) is given by:
$ M.P = \frac{L}{f_0} \left(1 + \frac{D}{f_e}\right), $
where
$L$ is the tube length of the microscope.
$f_0$ is the focal length of the objective lens.
$f_e$ is the focal length of the eyepiece.
$D$ is the least distance of distinct vision (usually taken as 250 mm, or 25 cm).
Step 2: Substitute the Known Values
We know:
$M.P = 375$
$L = 150\,\text{mm}$
$f_0 = 5\,\text{mm}$
$D = 250\,\text{mm}$
Substituting these values into the magnification formula:
$ 375 = \frac{150}{5} \left(1 + \frac{250}{f_e}\right).
$
Step 3: Simplify the Expression
First, calculate the factor $\frac{L}{f_0}$:
$ \frac{150}{5} = 30.
$
So the equation becomes:
$ 375 = 30 \left(1 + \frac{250}{f_e}\right).
$
Step 4: Solve for the Eyepiece Focal Length $f_e$
Divide both sides by 30:
$ 1 + \frac{250}{f_e} = \frac{375}{30} = 12.5.
$
Subtract 1 on both sides:
$ \frac{250}{f_e} = 12.5 - 1 = 11.5.
$
Finally, solve for $f_e$:
$ f_e = \frac{250}{11.5} \approx 21.7 \,\text{mm}.
$
This is approximately $22\,\text{mm}$.
Step 5: Conclusion
The required focal length of the eyepiece is about $22\,\text{mm}.$
