In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is __________

We are told that the compound microscope forms its final virtual image at the least distance of distinct vision, so we take $$D = 25\;\text{cm}.$$ The focal length of the objective lens is $$f_o = 1\;\text{cm}$$ and the distance between the two lenses (the tube length) is $$L = 20\;\text{cm}.$$ The required overall magnification is $$M = 100.$$ Our task is to find the focal length $$f_e$$ of the eye-piece lens.

For a compound microscope, the total magnification is the product of the objective’s lateral magnification and the eye-piece’s angular magnification: $$M = m_o \times M_e.$$

When the object is placed just beyond the first focal point of the objective, the real image it produces lies almost at the tube length. Under this commonly used approximation,

$$m_o = \frac{v_o}{u_o} \simeq \frac{L}{f_o}.$$

Substituting the given values,

$$m_o = \frac{20\;\text{cm}}{1\;\text{cm}} = 20.$$

The eye-piece behaves like a simple microscope. When its final image is at the near point $$D,$$ the angular magnification is

$$M_e = 1 + \frac{D}{f_e}.$$

Combining these two results in the expression for total magnification, we have

$$M = \left(\frac{L}{f_o}\right)\!\!\left(1 + \frac{D}{f_e}\right).$$

Now we insert the numerical data $$M = 100,\; L = 20\;\text{cm},\; f_o = 1\;\text{cm},\; D = 25\;\text{cm}:$$

$$100 = \left(\frac{20}{1}\right)\!\!\left(1 + \frac{25}{f_e}\right).$$

Dividing both sides by 20 gives

$$\frac{100}{20} = 1 + \frac{25}{f_e},$$ $$5 = 1 + \frac{25}{f_e}.$$

Subtracting 1 from both sides,

$$4 = \frac{25}{f_e}.$$

Finally, multiplying across,

$$f_e = \frac{25}{4}\;\text{cm} = 6.25\;\text{cm}.$$

So, the answer is $$6.25\;\text{cm}.$$