A small piece of wire bent into an L shape, with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10cm. I fthe bend is 20cm from the pole of the mirror, then the ration of the lengths of the images of the upright and horizontal portions of the wire is

Options

A. `1:2`
B. `3:1`
C. `1:3`
D. `2:1`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify Given Values
- Radius of curvature (R) of the concave mirror = 10 cm
- Object distance (u) from the pole of the mirror = -20 cm (the negative sign indicates that the object is in front of the mirror)
- Since the radius of curvature is given, we can find the focal length (f) using the formula:
\[ f = -\frac{R}{2} = -\frac{10}{2} = -5 \text{ cm} \]

### Step 2: Use the Mirror Formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{-5} = \frac{1}{v} + \frac{1}{-20} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-5} + \frac{1}{20} \]
Finding a common denominator (20):
\[ \frac{1}{v} = \frac{-4 + 1}{20} = \frac{-3}{20} \]
Thus,
\[ v = -\frac{20}{3} \text{ cm} \]

### Step 3: Calculate Magnification
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\left(-\frac{20}{3}\right) \div (-20) = \frac{20/3}{20} = \frac{1}{3} \]

### Step 4: Determine the Ratio of Lengths of Images
Since the lengths of the upright and horizontal portions of the wire are equal, we can denote them as L. The image of the upright portion will be magnified by the magnification factor, while the horizontal portion will be affected by the lateral magnification.

For the horizontal portion, we need to calculate the lateral magnification using the formula:
\[ m_{lateral} = \frac{f^2}{(u - f)^2} \]
Substituting the values:
\[ m_{lateral} = \frac{(-5)^2}{(-20 - (-5))^2} = \frac{25}{(-15)^2} = \frac{25}{225} = \frac{1}{9} \]

### Step 5: Calculate the Ratio of Lengths of Images
Let the length of the image of the upright portion be \(L_{upright}\) and the length of the image of the horizontal portion be \(L_{horizontal}\).
\[ L_{upright} = m \cdot L = \frac{1}{3}L \]
\[ L_{horizontal} = m_{lateral} \cdot L = \frac{1}{9}L \]
Now, the ratio of the lengths of the images of the upright and horizontal portions is:
\[ \text{Ratio} = \frac{L_{upright}}{L_{horizontal}} = \frac{\frac{1}{3}L}{\frac{1}{9}L} = \frac{1/3}{1/9} = \frac{9}{3} = 3 \]

### Final Answer
The ratio of the lengths of the images of the upright and horizontal portions of the wire is **3:1**.

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