A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such that its image is real and magnified and one end touches to rod. Its magnification will be

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

To solve the problem step by step, we will use the mirror formula and the magnification formula for concave mirrors.

### Step 1: Understand the given information
- Length of the rod (object height, h_o) = 5 cm
- Focal length (f) of the concave mirror = 10 cm
- The image is real and magnified, and one end of the image touches the rod.

### Step 2: Determine the radius of curvature (R)
The radius of curvature (R) is related to the focal length (f) by the formula:
\[ R = 2f \]
Substituting the given focal length:
\[ R = 2 \times 10 \, \text{cm} = 20 \, \text{cm} \]

### Step 3: Determine the object distance (u)
Since the image is real and magnified, the object must be placed beyond the center of curvature. The object distance (u) can be taken as:
\[ u = -R = -20 \, \text{cm} \]
(Note: The negative sign indicates that the object is placed in front of the mirror.)

### Step 4: Use the mirror formula to find the image distance (v)
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{-20} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{10} + \frac{1}{20} \]
Finding a common denominator (20):
\[ \frac{1}{v} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \]
Thus,
\[ v = \frac{20}{3} \, \text{cm} \]

### Step 5: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \]
Where:
- \( h_i \) is the height of the image
- \( h_o \) is the height of the object

Since the object height \( h_o = 5 \, \text{cm} \), we can calculate the magnification:
\[ m = -\frac{v}{u} = -\frac{\frac{20}{3}}{-20} = \frac{20/3}{20} = \frac{1}{3} \]

### Step 6: Determine the height of the image (h_i)
Using the magnification formula:
\[ h_i = m \cdot h_o = \frac{1}{3} \cdot 5 \, \text{cm} = \frac{5}{3} \, \text{cm} \]

### Step 7: Conclusion
The magnification is \( \frac{1}{3} \), which means the image is smaller than the object.

### Final Answer
The magnification will be \( \frac{1}{3} \).

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