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 at 1. 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= \|v₀\| + \|u_e\| ∴ v₀ = 15 – 5 = 10 \(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\) = \(\frac{1}{10}-\frac{1}{2}\) 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 at 1. 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= |v₀| + |u_e|

∴ v₀ = 15 – 5 = 10

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\) = \(\frac{1}{10}-\frac{1}{2}\)

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