Question

The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece is 5. When the microscope is focused on a certain object, the distance between the objective and eyepiece is observed to be 14 cm. Calculate the focal lengths of the objective and the eyepiece. (Given that the least distance of distinct vision = 25 cm)

Hint:

In compound microscopes, total magnification is the product of objective and eyepiece magnifications. Use \( M_e = 1 + \frac{D}{f_e} \) for near-point viewing and \( M_o = \frac{L - f_e}{f_o} \) for calculating the objective focal length.

Answer 1

Given: - Total magnification: \( M = 20 \) - Eyepiece magnification: \( M_e = 5 \) - Length of microscope: \( L = 14\,\text{cm} \) - Least distance of distinct vision: \( D = 25\,\text{cm} \) Step 1: Objective magnification \[ M = M_o \cdot M_e \Rightarrow M_o = \frac{M}{M_e} = \frac{20}{5} = 4 \] Step 2: Formula for objective magnification: \[ M_o = \frac{L - f_e}{f_o} \Rightarrow 4 = \frac{14 - f_e}{f_o} \quad \cdots (1) \] Step 3: Formula for eyepiece magnification (for relaxed eye): \[ M_e = 1 + \frac{D}{f_e} \Rightarrow 5 = 1 + \frac{25}{f_e} \Rightarrow \frac{25}{f_e} = 4 \Rightarrow f_e = \frac{25}{4} = 6.25\,\text{cm} \] Step 4: Substitute \( f_e \) in (1): \[ 4 = \frac{14 - 6.25}{f_o} \Rightarrow 4 = \frac{7.75}{f_o} \Rightarrow f_o = \frac{7.75}{4} = 1.9375\,\text{cm} \] % Final Answer Statement Answer: - Focal length of objective \( f_o = 1.9375\,\text{cm} \) - Focal length of eyepiece \( f_e = 6.25\,\text{cm} \)