Compound microscope – problems and solutions

1.

Compound microscope – problems and solutions 1

Based on the figure above, what is the overall magnification? Assume a normal eye so the near point (N) = 25 cm.

Known :

The focal length of the objective lens (fo) = 2 cm

The distance between the object and the objective lens (do) = 2.2 cm

The near point of the normal eye (N) = 25 cm

The focal length of the eyepiece lens (fe) = 25 cm

Wanted: The overall magnification (M)

Solution :

If the final image formed by the eyepiece lens at infinity then the eye is relaxed. If the final image not at infinity then the eye is not relaxed.

Based on the figure above, the distance of the virtual image formed by the eyepiece lens at infinity. The final image at infinity so that the eye is relaxed.

Compound microscope – problems and solutions 2

First, calculate the image distance from the objective lens (do‘). The objective lens is the converging lens so that the image distance calculated using the equation of the converging lens.

The focal length of the converging lens is positive.

1/do‘ = 1/fo – 1/do = 1/2 – 1 / 2.2 = 11 / 22 – 10/22 = 1/22

do‘ = 22/1 = 22 cm

The overall magnification of the microscope :

Compound microscope – problems and solutions 3

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2.

Compound microscope – problems and solutions 4

Based on the figure above, what is the distance between the objective lens and the eyepiece lens of the microscope?

Solution :

The microscope has both objective and eyepiece (ocular) lenses. A microscope is used to view objects that are very close. The image formed by the objective lens is the real image. The image is magnified by the eyepiece lens into a very large virtual image.

Known :

The focal length of the objective lens (fo) = 1.8 cm (The focal length is positive because the objective lens is the converging lens)

The focal length of the eyepiece lens (fe) = 6 cm (The focal length is positive because the ocular lens is the converging lens)

The distance between the object and the objective lens (do) = 2 cm (plus sign indicates that image is real)

Wanted: The distance between the objective lens and the eyepiece lens (length of microscope = d)

Solution :

The distance between the image and the objective lens (do’) :

1/do + 1/do’ = 1/fo

1/do’ = 1/fo – 1/do

1/do’ = 1 / 1.8 – 1/2 = 10/18 – 9/18 = 1/18

do’ = 18 cm

The real image formed by the objective lens located at the focal point of the eyepiece lens, as shown on the figure above.

The distance between the objective lens and the ocular lens :

l = do’ + fe = 18 cm + 6 cm = 24 cm

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3.

Compound microscope – problems and solutions 5

The distance between the object and the objective lens is 1.1 cm, the focal length of the objective lens is 1 cm and the focal length of the ocular lens is 5 cm. What is the overall magnification of the microscope?

Known :

The distance between the object and the objective lens (do) = 1.1 cm

The focal length of the objective lens (fo) = 1 cm (The focal length is positive because the objective lens is the converging lens)

The focal length of the ocular lens (fe) = 5 cm (The focal length is positive because the ocular lens is the converging lens)

Wanted: The overall magnification (M)

Solution :

The overall magnification of a microscope is the product of the magnification of the objective lens (mo) times the angular magnification (Me) of the eyepiece lens.

The magnification of the objective lens :

The equation of the linear magnification of the objective lens (mob) :

mo = ho’/ho = do’/do = (l – fe)/do

where

ho’ = the height of image formed by the objective lens

ho = the height of object

do’ = the distance between the real image formed by the objective lens and the objective lens

do = distance between object and the objective lens

fe = the focal length of the ocular lens

l = the length of microscope = distance between the two lenses = the distance between the real image formed by the objective lens (do’) + the focal length of the ocular lens fe)

The distance between the real image formed by the objective lens and the objective lens (do’) :

First, determine the distance between the real image and the objective lens :

1/do + 1/do’ = 1/fo

1/do’ = 1/fo – 1/do

1/do’ = 1/1 – 1 / 1.1 = 11/11 – 10/11 = 1/11

do’ = 11 cm

The real image formed by the objective lens is at the ocular focal point, as shown in the figure above.

The magnification of the objective lens :

mo = do’ / do = 11 cm / 1.1 cm = 10

The magnification of the ocular lens :

If the eye is relaxed, then the magnification of the ocular lens (Me) :

Me = N / fe

Where

N = the near point of eye (25 cm)

fe = the focal length of the ocular lens = 5 cm

The angular magnification of the ocular lens (Me) :

Me = N / fe = 25 cm / 5 cm = 5

The overall magnification of the microscope :

M = mo x Me = 10 x 5 = 50

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4.

Compound microscope – problems and solutions 6

The distance between the objective lens and the ocular lens of the microscope.

Known :

The distance between object and the objective lens (do) = 2 cm

The focal length of the objective lens (fo) = 1.8 cm

The distance between the real image and the ocular lens (de) = 6 cm

The focal length of the ocular lens (fe) = 6 cm

Wanted: The distance between the objective lens and the ocular lens (the length of the microscope)

Solution :

If the eye is relaxed, the final image formed by the ocular lens is at infinity. The final image is at infinity if the real image formed by the objective lens is at the focal point of the ocular lens.

The distance between the objective lens and the ocular lens (l) = the distance between the real image formed by the objective lens (do’) + the focal length of the ocular lens (fe).

The distance between the real image and the objective lens (do’) :

1/do + 1/do’ = 1/fo

1/do’ = 1/fo – 1/do

1/do’ = 1 / 1.8 – 1/2

1/do’ = 10/18 – 9/18 = 1/18

do’ = 18 cm

The distance between the objective lens and the ocular lens (the length of the microscope) :

l = do’ + fe

l = 18 cm + 6 cm

l = 24 cm