A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $$m_{1}$$ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $$m_{2}$$.
The value of $$ \mid\frac{m_{1}}{m_{2}}\mid $$ is______.

We have a thin convex lens of focal length $$f_1 = 5$$ cm and a thin concave lens of focal length $$f_2 = -4$$ cm. An object is placed 10 cm before the convex lens.

When the lenses are in contact with no gap, the combined focal length is given by

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} + \frac{1}{-4} = \frac{4 - 5}{20} = -\frac{1}{20}$$

So $$F = -20$$ cm (the combination acts as a diverging lens).

Applying the thin lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$$ with $$u = -10$$ cm yields

$$\frac{1}{v} = \frac{1}{F} + \frac{1}{u} = -\frac{1}{20} + \frac{1}{-10} = -\frac{1}{20} - \frac{2}{20} = -\frac{3}{20}$$

Thus $$v = -\frac{20}{3}$$ cm (virtual image on the same side as the object) and $$m_1 = \frac{v}{u} = \frac{-20/3}{-10} = \frac{2}{3}$$.

With a 1 cm gap between the lenses, the convex lens first produces an image. For the convex lens with $$u_1 = -10$$ cm and $$f_1 = 5$$ cm, the lens equation gives

$$\frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u_1} = \frac{1}{5} + \frac{1}{-10} = \frac{2 - 1}{10} = \frac{1}{10}$$

Hence $$v_1 = 10$$ cm (real image 10 cm to the right of the convex lens).

The image from the convex lens acts as a virtual object for the concave lens, which is placed 1 cm to its right. The distance from the concave lens to this image is $$10 - 1 = 9$$ cm on the far side, so $$u_2 = +9$$ cm. Applying the lens formula with $$f_2 = -4$$ cm yields

$$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-4} + \frac{1}{9} = \frac{-9 + 4}{36} = -\frac{5}{36}$$

Thus $$v_2 = -\frac{36}{5}$$ cm.

The total magnification is the product of the magnifications of each lens:

$$m_2 = \frac{v_1}{u_1} \times \frac{v_2}{u_2} = \frac{10}{-10} \times \frac{-36/5}{9} = (-1)\left(-\frac{4}{5}\right) = \frac{4}{5}$$

Therefore, the ratio of magnifications is

$$\left|\frac{m_1}{m_2}\right| = \left|\frac{2/3}{4/5}\right| = \left|\frac{2}{3} \times \frac{5}{4}\right| = \frac{10}{12} = \frac{5}{6}$$

Hence, $$\left|\frac{m_1}{m_2}\right| = \frac{5}{6}$$.