Question

The two lenses of a compound microscope are separated by a distance of 19.0 cm. If the objective lens produces a lateral magnification of 12.5✕ and the overall magnification is 110✕, determine the angular magnification of the eyepiece, the focal length of the eyepiece in cm, and the focal length of the objective lens in cm.

(a) the angular magnification of the eyepiece

(b) the focal length of the eyepiece in cm. ____cm

(c) the focal length of the objective lens in cm. ___cm

Answer #1

solution

a) overall magnification M = m1 X m2

linear magnification by objective = m1 = v/ u = image distance for objective lens/ object distance for objective distance

m1 = 12.5

M= 110

m2 = M/ m1 = 110/12.5 =8.8

(b)

angular magnification of eye piece m2 = D / f2

focal length of eyepiece f2 = D / m2 = 25 /8.8 = 2.84 cm

(c)

separation between objectice and eysepiece L = 19.0 cm

image distance for objective v = L - f2 = 19 - 2.84 = 16.16 cm

m1 = v / u

object distance u = v /m1 = 16.16 / 12.5 = 1.29 cm

lens formula 1/f = 1/v - 1/u

using sign convention u = -1.29 cm , v = + 16.16 cm

f= v u / (u - v)

ffocal length of objective lens f = 16.16 X - 1.29 / ( - 1.29- 16.16) = 1.2 cm

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