Step 1: Understanding the Concept:
A compound microscope consists of an objective lens and an eyepiece.
For a "relaxed eye" (normal adjustment), the final image is formed at infinity. This happens when the intermediate image formed by the objective lens falls exactly at the focal point of the eyepiece.
Step 2: Key Formula or Approach:
1. Tube length ($L$): For a relaxed eye, the distance between the lenses is $L = v_o + f_e$, where $v_o$ is the image distance for the objective and $f_e$ is the focal length of the eyepiece.
2. Magnifying Power ($M$): The total magnification for normal adjustment is $M = m_o \times m_e = \left(\frac{v_o}{u_o}\right) \times \left(\frac{D}{f_e}\right)$, where $u_o$ is the object distance and $D$ is the least distance of distinct vision (standard value $D = 25\text{ cm}$).
Step 3: Detailed Explanation:
Given values:
Length of microscope, $L = 15\text{ cm}$
Magnifying power, $M = 25$
Focal length of eyepiece, $f_e = 6\text{ cm}$
Least distance of distinct vision, $D = 25\text{ cm}$ (standard assumption)
First, find the image distance $v_o$ produced by the objective lens using the tube length formula:
\[ L = v_o + f_e \]
\[ 15 = v_o + 6 \]
\[ v_o = 15 - 6 = 9\text{ cm} \]
Next, use the magnifying power formula to find $u_o$:
\[ M = \left( \frac{v_o}{u_o} \right) \times \left( \frac{D}{f_e} \right) \]
Substitute the known values:
\[ 25 = \left( \frac{9}{u_o} \right) \times \left( \frac{25}{6} \right) \]
Divide both sides by 25:
\[ 1 = \left( \frac{9}{u_o} \right) \times \left( \frac{1}{6} \right) \]
\[ 1 = \frac{9}{6 \cdot u_o} \]
Simplify the fraction:
\[ 1 = \frac{3}{2 \cdot u_o} \]
Solve for $u_o$:
\[ u_o = \frac{3}{2} = 1.5\text{ cm} \]
Step 4: Final Answer:
The object distance for the objective lens is $1.5\text{ cm}$.