Question

Combination of lenses are arranged in case I and case II as shown in the figure. In case I, \( f_1 = 5 \, \text{cm} \), and in case II, \( f_2 = 4 \, \text{cm} \), the object is at \( -10 \, \text{cm} \). Magnification in two cases are \( m_1 \) and \( m_2 \). Find \( \left| \frac{m_1}{m_2} \right| \).

• A. \( \frac{5}{6} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{6}{5} \)

Hint:

When working with lenses, use the lens formula to find the image distance, and then calculate the magnification.

Answer 1

The Correct Option is A

To solve the problem, we need to find the magnification ratio \left| \frac{m_1}{m_2} \right| for the given lens arrangements in Cases I and II.

1. Case I:
   • Object distance u = -10 \, \text{cm} .
   • Focal length of the lens f_1 = 5 \, \text{cm} .
   • Using the lens formula: \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1}
   • Substitute the known values: \frac{1}{v} - \frac{1}{(-10)} = \frac{1}{5}
   • This simplifies to: \frac{1}{v} + \frac{1}{10} = \frac{1}{5}
   • Solve for v : \frac{1}{v} = \frac{1}{5} - \frac{1}{10} = \frac{1}{10}
   • Thus, v = 10 \, \text{cm} .
   • Magnification m_1 = \frac{v}{u} = \frac{10}{-10} = -1 .
2. Case II:
   • The first lens with focal length f_1 = 5 \, \text{cm} is the same as Case I.
   • The second lens with focal length f_2 = 4 \, \text{cm} at 1 \, \text{cm} from the first lens.
   • For the first lens: \frac{1}{v_1} - \frac{1}{(-10)} = \frac{1}{5}
   • This simplifies to: \frac{1}{v_1} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}
   • Thus, v_1 = \frac{10}{3} \, \text{cm} .
   • Magnification for the first lens m_1' = \frac{v_1}{u} = \frac{\frac{10}{3}}{-10} = -\frac{1}{3} .
   • The second lens: u_2 = v_1 - 1 = \frac{10}{3} - 1 = \frac{7}{3} \, \text{cm}
   • Using lens formula for the second lens: \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{4}
   • Substitute the values: \frac{1}{v_2} = \frac{1}{4} + \frac{3}{7} = \frac{19}{28}
   • This gives v_2 = \frac{28}{19} \, \text{cm} .
   • Magnification for the second lens m_2' = \frac{v_2}{u_2} = \frac{\frac{28}{19}}{\frac{7}{3}} = \frac{12}{19} .
   • Total magnification m_2 = m_1' \times m_2' = -\frac{1}{3} \times \frac{12}{19} = -\frac{4}{19} .
3. Ratio of magnifications: \left| \frac{m_1}{m_2} \right| = \left| \frac{-1}{-\frac{4}{19}} \right| = \frac{19}{4} = \frac{5}{6}

The correct answer is \frac{5}{6}.