An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to ___ cm.

In the first scenario, an object is viewed at the near point distance $$D = 25$$ cm using a single microscopic lens, giving a magnification of $$m_0 = 6$$. For a simple lens with the image at the near point, the magnification is $$m_0 = 1 + \frac{D}{f_{obj}}$$, giving $$6 = 1 + \frac{25}{f_{obj}}$$, so $$f_{obj} = 5$$ cm.

In the second scenario, a compound microscope is used with an eyepiece and a tube of length $$L = 0.6$$ m $$= 60$$ cm. The image is now observed at infinity (relaxed eye), and the total magnification is double the earlier, i.e., $$m_{total} = 12$$.

For a compound microscope with final image at infinity, the total magnification is:

where $$L$$ is the tube length (length of the microscope tube, taken as the image distance from the objective minus its focal length). Substituting $$f_{obj} = 5$$ cm, $$D = 25$$ cm, $$L = 60$$ cm, and $$m_{total} = 12$$:

Therefore, the focal length of the eyepiece is $$\boxed{25}$$ cm.