Question

. A very fine sample is placed 0.15 cm from the objective of a microscope. The focal length of the objective is 0.14 cm and of the eyepiece is 1.0 cm. The near-point distance of the person using the microscope is 25.0 cm. What is the final magnification of the microscope?

Answer 1

Answer:

Final magnification = -375

Explanation:

The total magnification of a compound microscope is expressed mathematically as the product of the magnifying power of each of the lenses that are combined in the compound microscope.

Final magnification = (magnifying power of the objective lens) × (magnifying power of the eyepiece lens) = m₁ × m₂

Magnifying power of a lens is defined as the ratio of the least distance of distinct vision to the focal length of the lens.

For the eyepiece, the least distance of distinct vision is the distance from the object to the near point = 25 + 1 + 0.15 = 26.15 cm

Focal length = 1 cm

Magnifying power of the eyepiece length = (26.15/1) = 26.15

For the objective lens,

The focal length = 0.14 cm

Least distance of distinct vision = -(1 + 1.00 + 0.15 - 0.14) = -2.01 cm (the negative sign means the image seen is upside down)

Magnifying power of the objective lens = (-2.01/0.14) = -14.357

Final magnification = -14.357 × 26.15 = -375

Hope this Helps!!!