A Lens Combination in a Microscope
A simple compound microscope uses an objective of focal length 1.0 cm and an eyepiece of focal length 5.0 cm separated by 20.0 cm. A specimen is placed 1.2 cm from the objective. The final image is viewed by a relaxed eye, so the eyepiece forms its image at infinity. A student claims the magnification is simply the product of the two focal lengths divided by the tube length. The objective and eyepiece are both converging lenses, and the setup is in air.
1. Locate the intermediate image formed by the objective and state whether it acts as a real object or a virtual object for the eyepiece.
Read the worked answer
Thin-lens equation for the objective: 1/f = 1/v − 1/u with sign convention (u = −1.2 cm, f = +1.0 cm).
1/v = 1/f + 1/u = 1/1.0 + 1/(−1.2) = 1 − 0.833 = 0.167 → v ≈ +6.0 cm.
The intermediate image is real, inverted, and 6.0 cm to the right of the objective.
Distance from objective to eyepiece is 20.0 cm, so the intermediate image lies 20.0 − 6.0 = 14.0 cm to the left of the eyepiece. It is a real object for the eyepiece.
Judgement: the sign convention must be applied consistently; a common error is to place the intermediate image beyond the eyepiece and call it virtual. Here it is comfortably in front of the eyepiece.
2. For the relaxed eye, find the eyepiece object distance required and check whether the given separation is consistent with a final image at infinity.
Read the worked answer
For the final image at infinity, the object for the eyepiece must be at its focal point: 5.0 cm in front of the eyepiece.
But the intermediate image is 14.0 cm in front of the eyepiece, not 5.0 cm. So with the stated 20.0 cm separation, the final image is not at infinity.
To achieve a relaxed eye, the eyepiece must be placed so that the intermediate image is 5.0 cm in front of it: separation = 6.0 + 5.0 = 11.0 cm.
Alternatively, keeping 20.0 cm, the object distance for the eyepiece is 14.0 cm, giving 1/v = 1/5.0 + 1/(−14.0) = 0.2 − 0.0714 = 0.129 → v ≈ +7.8 cm, a real image formed beyond the eyepiece, which the eye cannot view comfortably.
Judgement: the problem's stated geometry is inconsistent with relaxed viewing; the student must notice this rather than blindly computing magnification.
3. Using the corrected separation of 11.0 cm, compute the angular magnification and comment on the student's claim.
Read the worked answer
Magnification of objective: m₁ = v/u = 6.0 / 1.2 = 5 (in magnitude; image inverted).
For the eyepiece with the image at infinity, angular magnification is m₂ = D/f e = 25/5.0 = 5, where D = 25 cm is the near point.
Total angular magnification M = m₁ × m₂ = 5 × 5 = 25.
The student's claim (product of focal lengths divided by tube length) gives 1.0 × 5.0 / 20 = 0.25, which is dimensionally and numerically wrong. The correct idea is that the objective produces a linear magnification v/u, and the eyepiece adds an angular magnification D/f e.
Judgement: magnification is a product of a linear factor and an angular factor, not a ratio of focal lengths; the tube length affects only whether the intermediate image falls at the eyepiece focal point.