An angular magnification of 30 X is desired using an objective lens of focal length 1.25 cm and an eyepiece of focal length 5 cm. What should be the separation between objective and eyepiece when the final image is formed at the near point?

11.67 cm

The correct answer is Option (1) → 11.67 cm

Given:

$M = 30$, $f_o = 1.25\text{ cm}$, $f_e = 5\text{ cm}$, $D = 25\text{ cm}$

Step 1: Eyepiece magnification

$m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6$

Step 2: Objective magnification

$M = m_o \times m_e \Rightarrow 30 = m_o \times 6 \Rightarrow m_o = 5$

Step 3: Find $v_o$ (objective image distance)

$m_o = \frac{v_o}{u_o} \Rightarrow u_o = -\frac{v_o}{5}$

Lens formula:

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

$\frac{1}{1.25} = \frac{1}{v_o} - \frac{1}{(-v_o / 5)} = \frac{1}{v_o} + \frac{5}{v_o} = \frac{6}{v_o}$

$v_o = 6 \times 1.25 = 7.5\text{ cm}$

Step 4: Find $u_e$ (object distance for eyepiece)

Final image at near point $\Rightarrow v_e = -25\text{ cm}$

$\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$

$\frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e}$

$\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25}$

$|u_e| = \frac{25}{6} \approx 4.17\text{ cm}$

Step 5: Separation (tube length)

$L = v_o + |u_e| = 7.5 + 4.17 = 11.67\text{ cm}$