An object is put at a distance of 5cm from the first focus of a convex lens of focal length 10cm. If a real image is formed, its distance from the lens will be

Options

A. 15cm
B. 20cm
C. 25cm
D. 30cm

Explanation

To solve the problem, we will use the lens formula for a convex lens, which is given by:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance from the lens,
- \( u \) is the object distance from the lens.

### Step 1: Identify the given values
From the problem, we have:
- Focal length \( f = 10 \, \text{cm} \) (for a convex lens, this is positive),
- Object distance \( u = -5 \, \text{cm} \) (the object is placed on the same side as the incoming light, hence negative).

### Step 2: Substitute the values into the lens formula
Using the lens formula:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting the values of \( f \) and \( u \):

\[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-5} \]

### Step 3: Simplify the equation
This can be rewritten as:

\[ \frac{1}{10} = \frac{1}{v} + \frac{1}{5} \]

To combine the fractions on the right side, we need a common denominator. The common denominator between \( v \) and \( 5 \) is \( 5v \):

\[ \frac{1}{10} = \frac{5 + v}{5v} \]

### Step 4: Cross-multiply to solve for \( v \)
Cross-multiplying gives:

\[ 5v \cdot \frac{1}{10} = 5 + v \]

This simplifies to:

\[ \frac{5v}{10} = 5 + v \]

or

\[ \frac{v}{2} = 5 + v \]

### Step 5: Rearranging the equation
Rearranging gives:

\[ \frac{v}{2} - v = 5 \]

This can be rewritten as:

\[ -\frac{v}{2} = 5 \]

Multiplying both sides by -2 gives:

\[ v = -10 \]

### Step 6: Solve for \( v \)
Now, substituting back into the equation:

\[ v = 30 \, \text{cm} \]

### Conclusion
The distance of the real image from the lens is \( v = 30 \, \text{cm} \).