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

Options

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

Explanation

To solve the problem step by step, we will follow the principles of optics, particularly the behavior of concave mirrors and the magnification formula.

### Step 1: Identify the given data
- The radius of curvature (R) of the concave mirror is 10 cm.
- The object (L-shaped wire) has both upright and horizontal portions of equal lengths of 10 cm each.
- The distance from the pole of the mirror to the bend of the wire is 20 cm.

### Step 2: Calculate the focal length (f) of the mirror
The focal length (f) of a concave mirror is given by the formula:
\[ f = \frac{R}{2} \]
Substituting the value of R:
\[ f = \frac{10 \text{ cm}}{2} = 5 \text{ cm} \]

### Step 3: Determine the object distance (u)
The object distance (u) is taken as negative in mirror conventions. Since the distance from the pole to the bend is 20 cm, we have:
\[ u = -20 \text{ cm} \]

### Step 4: Use the mirror formula to find the image distance (v)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
Rearranging the formula to find \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
Substituting the values of f and u:
\[ \frac{1}{v} = \frac{1}{-5} - \frac{1}{-20} \]
\[ \frac{1}{v} = -\frac{1}{5} + \frac{1}{20} \]
Finding a common denominator (20):
\[ \frac{1}{v} = -\frac{4}{20} + \frac{1}{20} = -\frac{3}{20} \]
Thus:
\[ v = -\frac{20}{3} \text{ cm} \]

### Step 5: Calculate the magnification (m)
The linear magnification for mirrors is given by:
\[ m = -\frac{v}{u} \]
Substituting the values of v and u:
\[ m = -\left(-\frac{20}{3}\right) / (-20) \]
\[ m = \frac{20}{3} \times \frac{1}{20} = \frac{1}{3} \]

### Step 6: Calculate the lateral magnification (M)
The lateral magnification is given by:
\[ M = \frac{f^2}{(u - f)^2} \]
Substituting the values:
\[ M = \frac{5^2}{(20 - 5)^2} = \frac{25}{15^2} = \frac{25}{225} = \frac{1}{9} \]

### Step 7: Find the ratio of the lengths of the images of the upright and horizontal portions
The ratio of the lengths of the images of the upright and horizontal portions can be calculated as follows:
- The length of the image of the upright portion is given by \( m \) and the horizontal portion by \( M \).
- Thus, the ratio is:
\[ \text{Ratio} = \frac{m}{M} = \frac{\frac{1}{3}}{\frac{1}{9}} = \frac{1}{3} \times \frac{9}{1} = 3 \]

### Final Answer
The ratio of the lengths of the images of the upright and horizontal portions of the wire is:
\[ \text{Ratio} = 3:1 \]

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