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: \( M_e = 5 \) - Microscope length: \( L = 14\,\text{cm} \) - Least distance of distinct vision: \( D = 25\,\text{cm} \) Step 1: Calculate Objective Magnification The objective magnification \( M_o \) is derived from the total magnification \( M \) and the eyepiece magnification \( M_e \) using the formula \( M = M_o \cdot M_e \). \[ M_o = \frac{M}{M_e} = \frac{20}{5} = 4 \] Step 2: Objective Magnification Formula The formula for objective magnification is given by \( M_o = \frac{L - f_e}{f_o} \). Substituting the calculated \( M_o \): \[ 4 = \frac{14 - f_e}{f_o} \quad \cdots (1) \] Step 3: Eyepiece Magnification Formula (Relaxed Eye) For a relaxed eye, the eyepiece magnification \( M_e \) is calculated using \( M_e = 1 + \frac{D}{f_e} \). Solving for \( f_e \): \[ 5 = 1 + \frac{25}{f_e} \] \[ \frac{25}{f_e} = 5 - 1 = 4 \] \[ f_e = \frac{25}{4} = 6.25\,\text{cm} \] Step 4: Substitute \( f_e \) into Equation (1) Substitute the calculated value of \( f_e \) into equation (1) to find \( f_o \): \[ 4 = \frac{14 - 6.25}{f_o} \] \[ 4 = \frac{7.75}{f_o} \] \[ f_o = \frac{7.75}{4} = 1.9375\,\text{cm} \] % Final Answer Statement Answer: - Focal length of objective \( f_o = 1.9375\,\text{cm} \) - Focal length of eyepiece \( f_e = 6.25\,\text{cm} \)