Question

In a microscope, the objective has a focal length \(f_o=2\) cm and the eye-piece has a focal length \(f_e=4\) cm. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is_____.

Hint:

Total Magnification \(M = m_o \times m_e\). For normal adjustment, \(m_e = D/f_e\). Find \(m_o\) using \(v_o\) derived from the tube length.

Answer 1

Correct Answer: 24

Step 1: Understanding Normal Adjustment:
In normal adjustment, the final image is formed at infinity. This implies the image formed by the objective lies at the focal point of the eyepiece. Distance of image from objective \(v_o\). Distance of object from eyepiece \(u_e = f_e = 4\) cm. Step 2: Tube Length Relationship:
Tube length \(L_{tube}\) is the distance between the lenses. \[ L_{tube} = v_o + u_e \] \[ 32 = v_o + 4 \implies v_o = 28 \, \text{cm} \] Step 3: Objective Magnification (\(m_o\)):
Using lens formula: \(\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}\). \[ \frac{1}{28} - \frac{1}{u_o} = \frac{1}{2} \] \[ -\frac{1}{u_o} = \frac{1}{2} - \frac{1}{28} = \frac{14 - 1}{28} = \frac{13}{28} \] \[ u_o = -\frac{28}{13} \, \text{cm} \] Magnification \(m_o = \frac{v_o}{u_o} = \frac{28}{28/13} = 13\). (Taking magnitude). Step 4: Total Magnification:
\[ M = m_o \times m_e \] For normal adjustment, \(m_e = \frac{D}{f_e} = \frac{25}{4} = 6.25\). \[ M = 13 \times 6.25 = 81.25 \] Step 5: Final Answer:
81.25