Answer: (4.48)

Solution

According to question, final image i.e., $v_2 = 25$ cm, $f_0 = 1$ cm, magnification, $m = m_1 m_2 = 100$ Using lens formula, For first lens or objective: $\frac{1}{v_1} - \frac{1}{-x} = \frac{1}{1} \Rightarrow v_1 = \frac{x}{x-1}$ Also magnification $|m_1| = \left|\frac{v_1}{u_1}\right| = \frac{1}{x-1}$ For 2nd lens or eye-piece, this is acting as object $\therefore u_2 = -(20 - v_1) = -\left(20 - \frac{x}{x-1}\right)$ and $v_2 = -25$ cm Angular magnification $|m_2| = \left|\frac{D}{u_2}\right| = \left|\frac{25}{u_2}\right|$ Total magnification $m = m_1 m_2 = 100$ $\left(\frac{1}{x-1}\right)\left(\frac{25}{20 - \frac{x}{x-1}}\right) = 100$ $\Rightarrow \frac{25}{20(x-1) - x} = 100 \Rightarrow 1 = 80(x-1) - 4x$ $\Rightarrow 76x = 81 \Rightarrow x = \frac{81}{76}$ $\Rightarrow u_2 = \left(20 - \frac{76}{81-76}\right) = \frac{-19}{5}$ Again using lens formula for eye-piece $\frac{1}{-25} - \frac{1}{\frac{-19}{5}} = \frac{1}{f_e} \Rightarrow f_e = \frac{25 \times 19}{106} = 4.48$ cm