A rod of length 30 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is

Options

A. 10 cm
B. 15 cm
C. 2.5 cm
D. 7.5 cm

Answer

To solve the problem step by step, we will use the mirror formula and the concept of image formation by concave mirrors.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Length of the rod (L) = 30 cm
- Focal length of the concave mirror (F) = -10 cm (negative because it is a concave mirror)
- Distance of the nearest end of the rod from the mirror (U1) = -20 cm (negative as per sign convention)
- Distance of the far end of the rod from the mirror (U2) = U1 - Length of the rod = -20 cm - 30 cm = -50 cm

2. **Apply the Mirror Formula:**
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]

3. **Calculate Image Distance for the Nearest End (V1):**
- For U1 = -20 cm:
\[ \frac{1}{V1} = \frac{1}{-10} - \frac{1}{-20} \]
\[ \frac{1}{V1} = -\frac{1}{10} + \frac{1}{20} \]
Finding a common denominator (20):
\[ \frac{1}{V1} = -\frac{2}{20} + \frac{1}{20} = -\frac{1}{20} \]
Thus,
\[ V1 = -20 \text{ cm} \]

4. **Calculate Image Distance for the Far End (V2):**
- For U2 = -50 cm:
\[ \frac{1}{V2} = \frac{1}{-10} - \frac{1}{-50} \]
\[ \frac{1}{V2} = -\frac{1}{10} + \frac{1}{50} \]
Finding a common denominator (50):
\[ \frac{1}{V2} = -\frac{5}{50} + \frac{1}{50} = -\frac{4}{50} \]
Thus,
\[ V2 = -12.5 \text{ cm} \]

5. **Calculate the Length of the Image:**
The length of the image of the rod is the distance between the two image points (V1 and V2):
\[ \text{Length of the image} = |V1 - V2| = |-20 - (-12.5)| = |-20 + 12.5| = |-7.5| = 7.5 \text{ cm} \]

### Final Answer:
The length of the image of the rod is **7.5 cm**.

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