A convergent beam of light passes through a diverging lens of focal length `0.2 m` and comes to focus at a distance of `0.3 m` behind the lens. Find the position of the point at which the beam would converge in the absence of the lens.

Options

A. `0.12 m`
B. `0.6 m`
C. `0.3 m`
D. `0.15 m`

Explanation

To solve the problem step by step, we will use the lens formula and the information provided.

### Step 1: Identify the given values
- Focal length of the diverging lens (f) = -0.2 m (negative because it is a diverging lens)
- Image distance (v) = 0.3 m (behind the lens, hence positive)

### Step 2: Write the lens formula
The lens formula 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,
- \( u \) is the object distance (the distance of the object from the lens).

### Step 3: Rearrange the lens formula to find \( u \)
Rearranging the formula gives us:
\[ \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \]

### Step 4: Substitute the known values into the formula
Substituting \( v = 0.3 \, \text{m} \) and \( f = -0.2 \, \text{m} \):
\[ \frac{1}{u} = \frac{1}{0.3} - \frac{1}{-0.2} \]

### Step 5: Calculate the right-hand side
Calculating each term:
\[ \frac{1}{0.3} = \frac{10}{3} \quad \text{and} \quad \frac{1}{-0.2} = -5 \]
Thus,
\[ \frac{1}{u} = \frac{10}{3} + 5 \]
To combine these, convert 5 into a fraction with a common denominator:
\[ 5 = \frac{15}{3} \]
So,
\[ \frac{1}{u} = \frac{10}{3} + \frac{15}{3} = \frac{25}{3} \]

### Step 6: Solve for \( u \)
Now, take the reciprocal to find \( u \):
\[ u = \frac{3}{25} = 0.12 \, \text{m} \]

### Conclusion
The position of the point at which the beam would converge in the absence of the lens is \( 0.12 \, \text{m} \) in front of the lens.

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