Question

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 :

• A. $22\, mm$
• B. $12 \,mm$
• C. $33\, mm$
• D. $2 \,mm$

Answer 1

The Correct Option is A

To determine the focal length of the eyepiece needed for a compound microscope, we first use the formula for the total magnification of a compound microscope:

M = \left(\frac{L}{f_o}\right) \cdot \left(\frac{D}{f_e}\right)

where,

• M is the total magnification (375 in this case),
• L is the tube length of the microscope (150 mm),
• f_o is the focal length of the objective lens (5 mm),
• D is the least distance of distinct vision, commonly taken as 250 mm for a relaxed eye,
• f_e is the focal length of the eyepiece, which we need to find.

Rearranging the formula to solve for f_e, we get:

f_e = \frac{L \cdot D}{M \cdot f_o}

Substituting the given values:

f_e = \frac{150 \times 250}{375 \times 5}

Simplifying the expression, we have:

f_e = \frac{37500}{1875} = 20 mm approximately

However, due to the closeness of practical choices and rounding, this question intends to choose the closest available option,

Therefore, the focal length of the eyepiece should be close to 22\, mm.

Hence, the correct answer is 22\, mm.