A compound microscope in normal adjustment consists of an objective lens with a focal leng… — JAMB UTME

A compound microscope in normal adjustment consists of an objective lens with a focal length of $1.5\text{ cm}$ and an eyepiece with a focal length of $5.0\text{ cm}$ . An object is placed at a distance of $1.6\text{ cm}$ in front of the objective lens. Calculate the distance, in centimeters, between the objective lens and the eyepiece.

Answer: 29 cm

Answer

The distance between the objective lens and the eyepiece is 29.0 cm.

Applying the thin lens formula to the objective lens yields

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o}

, giving an image distance

v_o = 24.0\text{ cm}

. Under normal adjustment, the intermediate image falls on the focal point of the eyepiece, making

u_e = f_e = 5.0\text{ cm}

. The total separation between the two lenses is

L = v_o + u_e = 24.0 + 5.0 = 29.0\text{ cm}

Step-by-Step Solution

Apply the thin lens formula to the objective lens to find the intermediate image position

v_o

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o} \Rightarrow \frac{1}{v_o} = \frac{2}{3} - \frac{5}{8} = \frac{1}{24}\text{ cm}^{-1} \Rightarrow v_o = 24.0\text{ cm}

The objective lens forms a real, inverted, magnified image at distance

v_o

from the objective.

Identify the object distance for the eyepiece

u_e

under normal adjustment.

u_e = f_e = 5.0\text{ cm}

For normal adjustment of a optical instrument, the final image is formed at infinity, requiring the intermediate image to sit exactly at the principal focus of the eyepiece.

Sum the intermediate image distance and eyepiece object distance to obtain total lens separation

L

L = v_o + u_e = 24.0\text{ cm} + 5.0\text{ cm} = 29.0\text{ cm}

The separation of lenses in a compound microscope is the distance from the objective to the intermediate image plus the distance from the intermediate image to the eyepiece.

Key Concept

Compound microscope optics and lens separation in normal adjustment

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