Published by:
CGP EDU Academic Team
Published on: September 12, 2026
In a compound microscope, the magnified virtual image is formed at a distance of
from the eyepiece. The focal length of its objective lens is
. If the magnification is
and the tube length of the microscope is
, then the focal length of the eye-piece lens (in
) is
Text Solution
Verified by ExpertsThe correct answer is:
B
Step 1: In a compound microscope, the total magnification (M) can be expressed as the product of the magnification of the objective lens (M_o) and the eyepiece (M_e):
M = M_o * M_e.
Step 2: The focal length of the objective lens (f_o) is given, and we can also determine the tube length (L). Using the formula for a microscope, we have:
L = f_o + f_e, where f_e is the focal length of the eyepiece.
Step 3: By rearranging this formula, we find:
f_e = L - f_o.
Step 4: Given values are: L = 25 cm (tube length), f_o = 1 cm (focal length of objective). Substituting these values gives:
f_e = 25 cm - 1 cm = 24 cm.
Step 5: The magnification can also be calculated and compared if needed for confirmation, but as per the calculations, the correct answer from the options is B: 24 cm.
M = M_o * M_e.
Step 2: The focal length of the objective lens (f_o) is given, and we can also determine the tube length (L). Using the formula for a microscope, we have:
L = f_o + f_e, where f_e is the focal length of the eyepiece.
Step 3: By rearranging this formula, we find:
f_e = L - f_o.
Step 4: Given values are: L = 25 cm (tube length), f_o = 1 cm (focal length of objective). Substituting these values gives:
f_e = 25 cm - 1 cm = 24 cm.
Step 5: The magnification can also be calculated and compared if needed for confirmation, but as per the calculations, the correct answer from the options is B: 24 cm.
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