Question

A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is

• A. 150
• B. 250
• C. 100
• D. 125

Answer 1

Answer: (A)

150

Answer 2

The magnification of a compound microscope when the final image is formed at the distance of distinct vision is given by:

$M = \frac{L}{f_o} (1 + \frac{D}{f_e})$

Where:

• $L$ is the tube length
• $f_o$ is the focal length of the objective lens
• $f_e$ is the focal length of the eyepiece
• $D$ is the distance of distinct vision

Given values:

• $L = 40$ cm
• $f_o = 2$ cm
• $f_e = 4$ cm
• $D = 25$ cm

Substitute these values into the formula:

$M = \frac{40}{2} (1 + \frac{25}{4}) = 20 (1 + 6.25) = 20 (7.25) = 145$

The calculated value 145 is closest to 150 among the given options.