Two point objects are placed on principal axis of a thin converging lens. One is 20cm from the lens and other is on the other side of lens at a distance of 40 cm from the lens. The images of both objects coincide. The magnitude of focal length of lens is :-

Options

A. `80/3`cm
B. `40/3`cm
C. `40` cm
D. `20/3` cm

Explanation

To find the magnitude of the focal length of the lens, we will use the lens formula:

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

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

### Step 1: Identify the distances of the objects
- For the first object (Object 1), the distance from the lens is \( u_1 = -20 \, \text{cm} \) (negative because it is on the same side as the incoming light).
- For the second object (Object 2), the distance from the lens is \( u_2 = +40 \, \text{cm} \) (positive because it is on the opposite side of the lens).

### Step 2: Write the lens formula for both objects
For Object 1:
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} \]
Substituting \( u_1 = -20 \):
\[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{20} \]

For Object 2:
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting \( u_2 = 40 \):
\[ \frac{1}{f} = \frac{1}{v_2} - \frac{1}{40} \]

### Step 3: Set the image distances equal
Since the images of both objects coincide, we have \( v_1 = v_2 = v \). Therefore, we can set the two equations equal to each other:
\[ \frac{1}{v} + \frac{1}{20} = \frac{1}{v} - \frac{1}{40} \]

### Step 4: Simplify the equation
Subtract \( \frac{1}{v} \) from both sides:
\[ \frac{1}{20} = -\frac{1}{40} \]

### Step 5: Solve for \( v \)
Now, we can rearrange the equation:
\[ \frac{1}{20} + \frac{1}{40} = \frac{2}{40} + \frac{1}{40} = \frac{3}{40} \]
Thus, we have:
\[ \frac{3}{40} = \frac{2}{f} \]

### Step 6: Cross-multiply to find \( f \)
Cross-multiplying gives:
\[ 3f = 80 \]
So,
\[ f = \frac{80}{3} \, \text{cm} \]

### Conclusion
The magnitude of the focal length of the lens is:
\[ f = \frac{80}{3} \, \text{cm} \approx 26.67 \, \text{cm} \]