A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece.
For the compound microscope, we have:
Magnifying power, m = 100
Focal length of the objective, f0 = 0.5 cm Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
∴ vo + fe = 6.5 cm .....(1)
The magnifying power for normal adjustment is given by
m = `v_0/u_0 × D/f_e`
`=> 100 = -[1-v_0/0.5] × 25/f_e`
⇒ 2vo -4fe = 1 ..(2)
On solving equations (1) and (2), we get:
v0 =4.5 cm and fe =2 cm
Thus, the focal length of the eyepiece is 2 cm.
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