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 \(\left|\frac{m_1}{m_2}\right|\) is __________ .

Explanation:
We determine the ratio of magnitudes of magnification by analyzing the system under two distinct configurations:

Case 1: Lenses in direct contact with no gap (\( d = 0 \))
1. Equivalent Focal Length Calculation:
The focal length of the convex lens is \( f_1 = +5\text{ cm} \) and the focal length of the concave lens is \( f_2 = -4\text{ cm} \). When two thin lenses are in contact, the net focal length (\( F \)) of the combination is:
\( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \)
\( \frac{1}{F} = \frac{1}{5} + \frac{1}{-4} = \frac{4 - 5}{20} = -\frac{1}{20} \implies F = -20\text{ cm} \)

2. Image Distance and Magnification (\( m_1 \)):
The object is situated at a distance of \( u = -10\text{ cm} \) from the combined lens system. Using the standard lens formula:
\( \frac{1}{v_1} - \frac{1}{u} = \frac{1}{F} \)
\( \frac{1}{v_1} - \frac{1}{-10} = \frac{1}{-20} \implies \frac{1}{v_1} = -\frac{1}{20} - \frac{1}{10} = -\frac{3}{20} \implies v_1 = -\frac{20}{3}\text{ cm} \)
The linear magnification \( m_1 \) for this combined case is:
\( m_1 = \frac{v_1}{u} = \frac{-20/3}{-10} = \frac{2}{3} \)

Case 2: Lenses separated by a distance (\( d = 1\text{ cm} \))
1. Refraction through the First Lens (Convex Lens):
For the convex lens alone, the object distance is \( u_1 = -10\text{ cm} \) and \( f_1 = +5\text{ cm} \):
\( \frac{1}{v'_1} - \frac{1}{u_1} = \frac{1}{f_1} \)
\( \frac{1}{v'_1} - \frac{1}{-10} = \frac{1}{5} \implies \frac{1}{v'_1} = \frac{1}{5} - \frac{1}{10} = \frac{1}{10} \implies v'_1 = +10\text{ cm} \)
The magnification produced by the first lens is:
\( m'_1 = \frac{v'_1}{u_1} = \frac{10}{-10} = -1 \)

2. Refraction through the Second Lens (Concave Lens):
The real image formed by the convex lens at a distance of \( 10\text{ cm} \) serves as a virtual object for the shifted concave lens. Since the concave lens is separated from the convex lens by a distance of \( 1\text{ cm} \), the virtual object distance (\( u_2 \)) measured from the concave lens is:
\( u_2 = +10\text{ cm} - 1\text{ cm} = +9\text{ cm} \)
Using the lens formula with \( f_2 = -4\text{ cm} \):
\( \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \)
\( \frac{1}{v_2} - \frac{1}{9} = \frac{1}{-4} \implies \frac{1}{v_2} = \frac{1}{9} - \frac{1}{4} = \frac{4 - 9}{36} = -\frac{5}{36} \implies v_2 = -\frac{36}{5}\text{ cm} \)
The magnification produced by the second lens is:
\( m'_2 = \frac{v_2}{u_2} = \frac{-36/5}{9} = -\frac{4}{5} \)

3. Total Magnification (\( m_2 \)):
The net system magnification is the product of individual component magnifications:
\( m_2 = m'_1 \times m'_2 = (-1) \times \left(-\frac{4}{5}\right) = \frac{4}{5} \)

Finding the Ratio Magnitude:
Comparing the two calculated magnification parameters yields:
\( \left|\frac{m_1}{m_2}\right| = \frac{2/3}{4/5} = \frac{2}{3} \times \frac{5}{4} = \frac{5}{6} \)

Hence, Option A is the correct choice.