The magnifying power of a microscope with an objective of `5 mm` focal length is 400. The length of its tube is `20cm.` Then the focal length of the eye`-` piece is

Options

A. (a)200 cm
B. (b)160 cm
C. (c)2.5 cm
D. (d)0.1 cm

Explanation

To find the focal length of the eyepiece in a microscope given the magnifying power, we can use the formula for magnifying power (M) of a compound microscope:

\[ M = \frac{L}{f_o} \cdot \left(1 + \frac{D}{f_e}\right) \]

Where:
- \( M \) = magnifying power of the microscope
- \( L \) = length of the tube (distance between the objective and eyepiece)
- \( f_o \) = focal length of the objective
- \( D \) = near point distance (typically taken as 25 cm for a normal eye)
- \( f_e \) = focal length of the eyepiece

Given:
- \( M = 400 \)
- \( L = 20 \, \text{cm} = 200 \, \text{mm} \) (since the focal length of the objective is given in mm)
- \( f_o = 5 \, \text{mm} \)
- \( D = 25 \, \text{cm} = 250 \, \text{mm} \)

We can rearrange the formula to solve for \( f_e \):

1. Substitute the known values into the magnification formula:
\[ 400 = \frac{200}{5} \cdot \left(1 + \frac{250}{f_e}\right) \]

2. Calculate \( \frac{200}{5} \):
\[ \frac{200}{5} = 40 \]
So the equation becomes:
\[ 400 = 40 \cdot \left(1 + \frac{250}{f_e}\right) \]

3. Divide both sides by 40:
\[ 10 = 1 + \frac{250}{f_e} \]

4. Subtract 1 from both sides:
\[ 9 = \frac{250}{f_e} \]

5. Cross-multiply to solve for \( f_e \):
\[ 9f_e = 250 \]

6. Divide both sides by 9:
\[ f_e = \frac{250}{9} \approx 27.78 \, \text{mm} \]

Thus, the focal length of the eyepiece is approximately \( 27.78 \, \text{mm} \).