In a compound microscope an object is placed at a distance of L5 em from the objective of focal length 125 cm, If the eye-piece has a focal length of 5 em and the final image is formed at the near point, find the magnifying power of the microscope.

Question: In a compound microscope an object is placed at a distance of L5 em from the objective of focal length 125 cm, If the eye-piece has a focal length of 5 em and the final image is formed at the near point, find the magnifying power of the microscope.

The correct answer is –

The magnifying power of a compound microscope is given by the formula:

M = (angular magnification due to objective) x (angular magnification due to eyepiece)

where angular magnification due to objective is given by:

M_o = (25 cm)/(f_o)

and angular magnification due to eyepiece is given by:

M_e = (f_e)/(25 cm)

where f_o and f_e are the focal lengths of the objective and eyepiece, respectively.

First, let’s calculate the angular magnification due to the objective:

M_o = (25 cm)/(f_o) = (25 cm)/(125 cm) = 0.2

Next, let’s calculate the angular magnification due to the eyepiece:

M_e = (f_e)/(25 cm) = (5 cm)/(25 cm) = 0.2

Now, we can calculate the magnifying power of the microscope:

M = (angular magnification due to objective) x (angular magnification due to eyepiece) = M_o x M_e = 0.2 x 0.2 = 0.04

Therefore, the magnifying power of the microscope is 0.04.