A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. what is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e, when the final image is at infinity ),
(b) The final image is formed at the least distance of distinct vision (25 cm )

Options

A. 33.6
B. 66.12
C. 22.6
D. 11.6

Explanation

To solve the problem, we will calculate the magnifying power of the telescope under two different conditions: (a) when the telescope is in normal adjustment and (b) when the final image is formed at the least distance of distinct vision.

### Given Data:
- Focal length of the objective lens, \( f_o = 140 \, \text{cm} \)
- Focal length of the eyepiece, \( f_e = 5 \, \text{cm} \)
- Least distance of distinct vision, \( D = 25 \, \text{cm} \)

### (a) Magnifying Power in Normal Adjustment

When the telescope is in normal adjustment, the magnifying power \( M \) is given by the formula:

\[ M = \frac{f_o}{f_e} \]

Substituting the given values:

\[ M = \frac{140 \, \text{cm}}{5 \, \text{cm}} = 28 \]

Thus, the magnifying power of the telescope when in normal adjustment is **28**.

### (b) Magnifying Power when the Final Image is at Least Distance of Distinct Vision

When the final image is formed at the least distance of distinct vision, the magnifying power \( M \) is given by the formula:

\[ M = \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right) \]

Substituting the values into the formula:

\[ M = \frac{140 \, \text{cm}}{5 \, \text{cm}} \left( 1 + \frac{5 \, \text{cm}}{25 \, \text{cm}} \right) \]

Calculating the term inside the parentheses:

\[ 1 + \frac{5}{25} = 1 + 0.2 = 1.2 \]

Now substituting this back into the magnifying power formula:

\[ M = 28 \times 1.2 = 33.6 \]

Thus, the magnifying power of the telescope when the final image is formed at the least distance of distinct vision is **33.6**.

### Summary of Results:
- (a) Magnifying power in normal adjustment: **28**
- (b) Magnifying power when the final image is at least distance of distinct vision: **33.6**