Question

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Answer 1

d = 25 cm, f₀ = 8 mm = 0.8 cm

f_e = 2.5 cm, u₀ = – 9 mm = – 0.9cm.

1/u_e = 1/v_e - 1/f_e = 1/-25 - 1/2.5 = (-1 - 10)/25 = -11/25

∴ u_e = - 25/11 = - 2.27cm

1/v₀ = 1/u₀ + 1/f₀ = 1/-0.9 + 1/0.8 = (-0.8 + 0.9)/0.72 = 0.1/0.72

∴ v₀ = 0.72/0.1 = 7.2cm. eperation between two lens = 2.27 + 7.2 = 9.47 cm.

magnifying power, m = v₀/|u|(1 + d/f_e) = 7.2/0.9(1 + 25/2.5) = 88