Question

1. The focal lengths of the objective and the eyepiece of a microscope are 0.50 cm and 2.0 cm, respectively, and their separation adjusted for minimum eyestrain (with the final image at the viewer's far point) is 6.0 cm. The near point of the person using the microscope is 25 cm and the far point is infinity.

a. If the microscope is focused on a small object, what is the distance between the object and the objective lens?

b. If the microscope is focused on a small object, what is its final magnification?

2. The distance between the object and the eyepiece of a compound microscope is 25.0 cm. The focal length of its objective lens is 0.200 cm and the eyepiece has a focal length of 2.60 cm. A person with a near point of 25.0 cm and a far point at infinity is using the microscope.

a. What is the angular magnification obtainable using the eyepiece alone as a magnifying lens if the final image is at the person's far point?

b. What is the total magnification of the microscope when used by the person of normal eyesight?

Answer #1

1]

a] Image distance = v = - 25 cm

focal length of the eyepiece = 2 cm

Use the lens formula, 1/f = 1/v + 1/u

=> 1/2 = - 1/25 + 1/u

=> u = 1.852 cm

this is the object distance from the eyepiece.

Since the separation is adjusted for minimum eyestrain, the image distance for the objective lens will then be: v' = 6 - 1.852 = 4.148 cm

f ' = 0.5 cm

1/0.5 = 1/4.148 + 1/u

=> u = 0.568 cm

this is the distance between the physical object and objective lens.

b] Magnification = m = - (vv'/ff ')

=> m = - (25 x 4.148/(0.5 x 2)) = - 103.7.

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