A thin linear object of size 1mm is kept along the principal axis of a convex lens of focal length 10cm. The object is at 15cm from the lens. The length of the image is:

Options

A. 1mm
B. 4mm
C. 2mm
D. 8mm

Explanation

To find the length of the image formed by a convex lens when a thin linear object is placed along its principal axis, we can follow these steps:

### Step 1: Identify the given parameters
- Size of the object (h₀) = 1 mm = 0.1 cm (since we need to work in cm)
- Focal length of the lens (f) = 10 cm
- Object distance (u) = -15 cm (the object distance is taken as negative in lens formula)

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- f = focal length
- v = image distance
- u = object distance

### Step 3: Substitute the known values into the lens formula
Substituting f = 10 cm and u = -15 cm into the lens formula:
\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \]
This simplifies to:
\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \]

### Step 4: Solve for v
Rearranging the equation:
\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \]
Finding a common denominator (which is 30):
\[ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]
Thus, we have:
\[ v = 30 \text{ cm} \]

### Step 5: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = -\frac{v}{u} \]
Substituting the values of v and u:
\[ m = -\frac{30}{-15} = 2 \]

### Step 6: Calculate the size of the image (h₁)
The size of the image (h₁) can be calculated using the magnification:
\[ h₁ = m \cdot h₀ \]
Substituting the values:
\[ h₁ = 2 \cdot 0.1 \text{ cm} = 0.2 \text{ cm} = 2 \text{ mm} \]

### Step 7: Conclusion
The length of the image is 2 mm.

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