Question

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 the convex lens and the object undisturbed, a gap of \( 1 \) cm is introduced between the lenses by moving the concave lens away. This leads to a change in magnification of the total lens system to \( m_2 \).

The value of \( \dfrac{m_1}{m_2} \) is

• A. $\dfrac{5}{9}$
• B. $\dfrac{3}{2}$
• C. $\dfrac{5}{27}$
• D. $\dfrac{25}{27}$

Hint:

When lenses are separated, treat image of the first lens as the object for the second lens and calculate magnifications stepwise.

Answer 1

The Correct Option is A

To solve this problem, we need to analyze the behavior of the lens system in two scenarios and calculate the ratio of the magnifications \( \dfrac{m_1}{m_2} \).

Scenario 1: Lenses Combined (No Gap)

1. The focal length of the convex lens \( f_1 = +5 \) cm.
2. The focal length of the concave lens \( f_2 = -4 \) cm.
3. The object distance from the convex lens \( u = -10 \) cm (using the sign convention for lenses).
4. The effective focal length \( F \) for the combination of the two lenses is given by: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} + \frac{1}{-4} = \frac{1}{5} - \frac{1}{4} = \frac{4 - 5}{20} = -\frac{1}{20} \] \]
5. The effective focal length is \( F = -20 \) cm.
6. The magnification \( m_1 \) for this lens combination is given by the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \] where \( v \) is the image distance. Thus, \[ \frac{1}{v} = \frac{1}{F} + \frac{1}{u} = -\frac{1}{20} - \frac{1}{10} = -\frac{1}{20} - \frac{2}{20} = -\frac{3}{20} \] \] Hence, \( v = -\frac{20}{3} \) cm.
7. The magnification \( m_1 \) is: \[ m_1 = -\frac{v}{u} = -\left(\frac{-20/3}{-10}\right) = \frac{20}{30} = \frac{2}{3} \] \]

Scenario 2: Gap of 1 cm Introduced

1. With the gap, the lenses do not act as a simple combination anymore. First, consider the convex lens alone.
2. Using the lens formula with \( u = -10 \) cm and \( f_1 = +5 \) cm: \[ \frac{1}{v_1} = \frac{1}{f_1} + \frac{1}{u} = \frac{1}{5} - \frac{1}{10} = \frac{2 - 1}{10} = \frac{1}{10} \] \] Hence, \( v_1 = 10 \) cm.
3. The formed image is now \( 10 \) cm behind the convex lens and serves as the object for the concave lens.
4. The object distance for the concave lens is \( u_2 = 10 - 1 = 9 \) cm (since there is a gap of 1 cm).
5. Using \( f_2 = -4 \) cm for the concave lens: \[ \frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = -\frac{1}{4} + \frac{1}{-9} = -\frac{9+4}{36} = -\frac{13}{36} \] \] Hence, \( v_2 = -\frac{36}{13} \) cm.
6. The total optical path from the object to the final image gives the new image distance \( v' = 10 - \frac{36}{13} = \frac{94}{13} \) cm.
7. The magnification \( m_2 \) for the total system is: \[ m_2 = \left(-\frac{v_1}{u}\right) \cdot \left(-\frac{v_2}{u_2}\right) = \frac{10}{10} \cdot \frac{36}{117} = \frac{36}{117} = \frac{12}{39} = \frac{4}{13} \] \]

Calculating the Ratio \( \dfrac{m_1}{m_2} \)

1. Using the values for \( m_1 \) and \( m_2 \): \[ \frac{m_1}{m_2} = \frac{\frac{2}{3}}{\frac{4}{13}} = \frac{2}{3} \times \frac{13}{4} = \frac{2 \times 13}{3 \times 4} = \frac{26}{12} = \frac{13}{6} = \frac{5}{9} \] \]

Therefore, the correct answer is \(\frac{5}{9}\).