A compound microscope consists of an objective lens of focal length 2

1.	A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at1. Least distance of distinct vision.2. infinity
Answer» 1. v_e = -25 cm \(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\) = \(\frac{1}{-25}-\frac{1}{6.25}\) ∴ u₀ = 5 cm Length of the tube, L= \|v_o\| + \|u_e\| ∴ v_o = 15 - 5 = 10 \(\frac{1}{u_o}=\frac{1}{v_0}-\frac{1}{f_0}\) = \(\frac{1}{10}-\frac{1}{2}\) u_e = -2.5 cm 2. ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75 \(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\) = \(\frac{1}{8.75}\) - \(\frac{1}{2}\) u₀ = -2.59 cm

A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at1. Least distance of distinct vision.2. infinity

Answer»

1. v_e = -25 cm

\(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\) = \(\frac{1}{-25}-\frac{1}{6.25}\)

∴ u₀ = 5 cm

Length of the tube, L= |v_o| + |u_e|

∴ v_o = 15 - 5 = 10

\(\frac{1}{u_o}=\frac{1}{v_0}-\frac{1}{f_0}\)

= \(\frac{1}{10}-\frac{1}{2}\)

u_e = -2.5 cm

2. ∴ v₀ = 15 – f_e = 15 – 6.25 = 8.75

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\)

= \(\frac{1}{8.75}\) - \(\frac{1}{2}\)

u₀ = -2.59 cm