Question

In the normal adjustment of an astronomical telescope, the objective and eyepiece are 36 cm apart. If the magnifying power of the telescope is 8, find the focal lengths of the objective and eyepiece.

• A. \( F_o = 28 \, cm,\; F_e = 7 \, cm \)
• B. \( F_o = 28 \, cm,\; F_e = 4 \, cm \)
• C. \( F_o = 32 \, cm,\; F_e = 4 \, cm \)
• D. \( F_o = 4 \, cm,\; F_e = 32 \, cm \)

Hint:

For an astronomical telescope in normal adjustment, use \( F_o + F_e = \text{tube length} \) and \( M = \frac{F_o}{F_e} \).

Answer 1

The Correct Option is C

Step 1: Understand normal adjustment.
In normal adjustment of an astronomical telescope, the final image is formed at infinity.
In this condition, the distance between the objective lens and eyepiece is equal to the sum of their focal lengths.
\[ F_o + F_e = 36 \]

Step 2: Write the formula for magnifying power.
For an astronomical telescope in normal adjustment, magnifying power is given by:
\[ M = \frac{F_o}{F_e} \]

Step 3: Substitute the given magnifying power.
Given:
\[ M = 8 \]
So:
\[ \frac{F_o}{F_e} = 8 \]
\[ F_o = 8F_e \]

Step 4: Substitute in the length relation.
We know:
\[ F_o + F_e = 36 \]
Substituting \( F_o = 8F_e \):
\[ 8F_e + F_e = 36 \]

Step 5: Solve for eyepiece focal length.
\[ 9F_e = 36 \]
\[ F_e = \frac{36}{9} \]
\[ F_e = 4 \, cm \]

Step 6: Find objective focal length.
\[ F_o = 8F_e \]
\[ F_o = 8 \times 4 \]
\[ F_o = 32 \, cm \]

Step 7: Conclusion.
Thus, the focal length of the objective is \( 32 \, cm \), and the focal length of the eyepiece is \( 4 \, cm \).
\[ \boxed{F_o = 32 \, cm,\; F_e = 4 \, cm} \]