The distance between the eyepiece and the objective lens in a certain compound m

Question

The distance between the eyepiece and the objective lens in a certain compound microscope is 23.4 cm. The focal length of the objective is 0.560 cm, and that of the eyepiece is 1.50 cm. Find the overall magnification of the microscope. (The near point of the eye is 25 cm. Assume that the object is placed at the focal point of the objective lens, and one places the eyepiece at the near point of the eye.) -635.12 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

Explanation / Answer

The distance between the objective and eye piece lens in the compound microscope is

L = 23.4 cm

The magnification due to the objective lens is

m0 = L / f0

Here, focal length of the objective is f0.

Substitute 23.4 cm for L and 0.560 cm for f0 in the above equation,

m0 = -L / f0

     = -23.4 cm / 0.560 cm

     = -41.786

Now, the magnification due to the eyepiece lens is

me = 25 cm / fe

Here, focal length of the eyepiece lens is fe.

Substitute 1.50 cm for fe in the above equation,

me = 25 cm / fe

      = 25 cm / 1.50 cm

      = 16.667

The magnification of the compound microscope is equal to magnification due to objective lens times the magnification due to eyepiece lens.

Thus,

m = m0me

Substitute the values in the above equation,

m = m0me

    = (-41.786)(16.667)

    =- 696.45

Negative sign can be ignored for finding the magnitude of the magnification as it indicates the final image tobe inverted and real image.

Rounding off to three significant figures, the magnification of the compound microscope is 696.

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