Question

The focal lengths of the objective and the eyepiece of a compound microscope are 1 cm and 2 cm respectively. If the tube length of the microscope is 10 cm, the magnification obtained by the microscope for most suitable viewing by relaxed eye is:

• A. 250
• B. 200
• C. 150
• D. 125 \bigskip

Hint:

The total magnification of a compound microscope is the product of the magnification due to the objective and the eyepiece. For relaxed eye viewing, the formula \( M = \frac{D}{F_o} \times \frac{L}{F_e} \) gives the magnification.

Answer 1

The Correct Option is D

The magnification \( M \) of a compound microscope is given by the formula: \[ M = \frac{D}{F_o} \times \frac{L}{F_e} \] where: - \( D \) is the least distance of distinct vision (usually taken as 25 cm), - \( F_o \) is the focal length of the objective lens, - \( L \) is the tube length of the microscope, - \( F_e \) is the focal length of the eyepiece. Given: - \( D = 25 \, \text{cm} \), - \( F_o = 1 \, \text{cm} \), - \( F_e = 2 \, \text{cm} \), - \( L = 10 \, \text{cm} \). Substituting these values into the formula: \[ M = \frac{25}{1} \times \frac{10}{2} = 25 \times 5 = 125 \] Thus, the magnification obtained by the microscope is 125. \bigskip