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. Real, inverted; Magnification \( = -50 \)
• B. Virtual, inverted; Magnification \( = 30 \)
• C. Virtual, erect; Magnification \( = 50 \)
• D. Real, inverted; Magnification \( = -30 \)

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Understand the microscope.
The objective lens makes a real, inverted, enlarged image of the object. This image then becomes the object for the eyepiece. The total magnification is the product of the two magnifications.
Step 2: Find the image distance for the objective.
With $f_o = 1$ cm and $u_o = -1.2$ cm, use the lens formula: \[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \;\Rightarrow\; 1 = \frac{1}{v_o} + \frac{1}{1.2} \]
Step 3: Solve for the objective image.
\[ \frac{1}{v_o} = 1 - \frac{5}{6} = \frac{1}{6} \;\Rightarrow\; v_o = 6\ \text{cm} \] Since $v_o$ is positive, the intermediate image is real and inverted.
Step 4: Find the objective magnification.
\[ m_o = \frac{v_o}{u_o} = \frac{6}{-1.2} = -5 \]
Step 5: Find the eyepiece magnification.
The final image is at the near point $D = 25$ cm, so the eyepiece magnification is: \[ m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6 \]
Step 6: Multiply for total magnification.
\[ m = m_o \times m_e = (-5)(6) = -30 \] So the image is real and inverted with total magnification: \[ \boxed{m = -30} \]