An optical instrument employed for angular magnification contains...

Given the power of the objective lens, $P_o = 25 \text{ D}$, its focal length is:

$f_o = \dfrac{100}{P_o} = \dfrac{100}{25} = 4 \text{ cm}$

Given the power of the eyepiece, $P_e = 20 \text{ D}$, its focal length is:

$f_e = \dfrac{100}{P_e} = \dfrac{100}{20} = 5 \text{ cm}$

For the eye to be least strained, the final image must be formed at infinity. This requires the intermediate image formed by the objective to lie exactly at the first focal point of the eyepiece. Thus, the object distance for the eyepiece is:

$u_e = f_e = 5 \text{ cm}$

The tube length $L$ of the microscope is the distance between the objective and the eyepiece:

$L = v_o + u_e$

$25 = v_o + 5 \Rightarrow v_o = 20 \text{ cm}$

Using the lens formula for the objective to find the object distance $u_o$:

$\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$

$\dfrac{1}{20} - \dfrac{1}{u_o} = \dfrac{1}{4}$

$\dfrac{1}{u_o} = \dfrac{1}{20} - \dfrac{1}{4} = -\dfrac{4}{20} = -\dfrac{1}{5} \text{ cm}^{-1}$

$u_o = -5 \text{ cm}$

The linear magnification produced by the objective is:

$m_o = \left| \dfrac{v_o}{u_o} \right| = \left| \dfrac{20}{-5} \right| = 4$

The angular magnification produced by the eyepiece (acting as a simple microscope focused at infinity) is:

$m_e = \dfrac{D}{f_e} = \dfrac{25}{5} = 5$

The total angular magnification of the optical instrument is:

$M = m_o \times m_e = 4 \times 5 = 20$

Answer: $20$