Problem 25.45 7 of 14 Constants Part A The eyepiece of a compound microscope has

Question

Problem 25.45 7 of 14 Constants Part A The eyepiece of a compound microscope has a focal length of 3.10 em, and the objective lens hasf 0.720 c If an object is placed 0.780c from the objective lens, calculate the distance between the lenses when the microscope is adjusted for a relaxed eye. Express your answer using two significant figures. Subnit Request Answer Part B Calculate the total magnification. Express your answer using two significant figures. Submit Request Answer Provide Feedback Next>

Explanation / Answer

A.

Using lens equation

1/f = 1/u + 1/v

u = object distance = 0.78 cm

f = focal kength = 0.72 cm

v = image distance = ?

v = f*u/(u - f)

v = 0.72*0.78/(0.78 - 0.72)

v = 9.36 cm

Now distance between the lenses will be

L = image distance + focal length of eyepiece

L = 9.36 cm + 3.10 cm

L = 12.46 cm

B.

Magnification will be given by:

M = Me*mo = (N/fe)*(v/u)

M = (25/3.10)*(9.36/0.78)

M = 96.77

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