Question:
A compound microscope consists of an objective lens of focal length \( f_{o}=1~\text{cm} \) and an eyepiece of focal length \( f_{e}=5~\text{cm} \). An object is placed 1.2 cm in front of the objective. The final image is formed at the least distance of distinct vision. The nature of the intermediate image and the total magnification of the microscope are:
A compound microscope consists of an objective lens of focal length \( f_{o}=1~\text{cm} \) and an eyepiece of focal length \( f_{e}=5~\text{cm} \). An object is placed 1.2 cm in front of the objective. The final image is formed at the least distance of distinct vision. The nature of the intermediate image and the total magnification of the microscope are:
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The total magnification sign in optical instruments like microscopes or telescopes is negative because the final image is inverted relative to the original object direction, which easily narrows down your choice parameters.
Updated On: Jun 8, 2026
- Real, inverted; Magnification \( = -50 \)
- Virtual, inverted; Magnification \( = 30 \)
- Virtual, erect; Magnification \( = 50 \)
- Real, inverted; Magnification \( = -30 \)
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The Correct Option is D
Solution and Explanation
Concept:
In a compound microscope, the objective lens forms a real, inverted, and magnified intermediate image of the object. This intermediate image then acts as an object for the eyepiece lens.
When the final image is formed at the least distance of distinct vision (\( D = 25~\text{cm} \)), the total magnifying power \( m \) is given by:
\[
m = m_o \times m_e = \left( \frac{v_o}{u_o} \right) \left( 1 + \frac{D}{f_e} \right)
\]
Step 1: Finding image position \( v_o \) for the objective lens.
Given parameters for the objective: \( f_o = 1~\text{cm} \), \( u_o = -1.2~\text{cm} \). Using the lens equation: \[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{1} = \frac{1}{v_o} - \frac{1}{-1.2} \] \[ 1 = \frac{1}{v_o} + \frac{1}{1.2} \implies \frac{1}{v_o} = 1 - \frac{1}{1.2} = 1 - \frac{5}{6} = \frac{1}{6} \implies v_o = 6~\text{cm} \] Since \( v_o \) is positive, the intermediate image is real and inverted.
Step 2: Calculating objective magnification \( m_o \).
\[ m_o = \frac{v_o}{u_o} = \frac{6}{-1.2} = -5 \]
Step 3: Calculating eyepiece magnification \( m_e \) and total magnification \( m \).
Given eyepiece parameter \( f_e = 5~\text{cm} \), and near point \( D = 25~\text{cm} \): \[ m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 1 + 5 = 6 \] Now, compute the total magnification: \[ m = m_o \times m_e = (-5) \times 6 = -30 \]
Step 1: Finding image position \( v_o \) for the objective lens.
Given parameters for the objective: \( f_o = 1~\text{cm} \), \( u_o = -1.2~\text{cm} \). Using the lens equation: \[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \implies \frac{1}{1} = \frac{1}{v_o} - \frac{1}{-1.2} \] \[ 1 = \frac{1}{v_o} + \frac{1}{1.2} \implies \frac{1}{v_o} = 1 - \frac{1}{1.2} = 1 - \frac{5}{6} = \frac{1}{6} \implies v_o = 6~\text{cm} \] Since \( v_o \) is positive, the intermediate image is real and inverted.
Step 2: Calculating objective magnification \( m_o \).
\[ m_o = \frac{v_o}{u_o} = \frac{6}{-1.2} = -5 \]
Step 3: Calculating eyepiece magnification \( m_e \) and total magnification \( m \).
Given eyepiece parameter \( f_e = 5~\text{cm} \), and near point \( D = 25~\text{cm} \): \[ m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 1 + 5 = 6 \] Now, compute the total magnification: \[ m = m_o \times m_e = (-5) \times 6 = -30 \]
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