A compound microscope consists of two converging lenses. Lens 1, called the obje

Question

A compound microscope consists of two converging lenses. Lens 1, called the objective, is a distance 61.0 mm from lens 2, called the eyepiece. The objective is placed in between the object and the eyepiece. The objective has a focal length of 15.0 mm, and the eyepiece has a focal length of 25.5 mm. If the microscope is used to examine an object placed 24.1 mm in front of the objective,
a) What is the distance of the final image with respect to the eyepiece?
b) Is the image real or virtual?
c) Is the image inverted or non-inverted?
d) What is the overall magnification of the microscope?

Explanation / Answer

focal length of object lens fo = 15mm = 1.5cm

focal length of eye piece fe =25.5mm = 2.55cm

distance between lens L = 61mm = 6.1cm

least distance of distant vision D = 25cm

d) total magnification m = LD/fofe = (6.1*25)/(1.5*2.55)

m =152.5/3.825 = 39.8

c) the final image is inverted

b) the final image is virtual

a) mo is magnification of object lens and me is magnification of eye piece then m = mo*me

where as objective lens behaves as simple microscope with magnification mo = D/fo = 25/1.5 = 16.67

magnification of eye piece me = m/mo = 39.8/16.67 = 2.39

me = v/u implies u = v/me = v/2.39

from (1/f) = (1/u)+(1/v)

f = (uv)/(u+v)

2.55 = ((v/2.39)*v)/((v/2.39)+v)

solving

v = 8.64cm is the distance of final image from eye piece

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