Lenses of powers 3D and -5D are combined to from a compound lens. An object is placed at a distance of 50 cm from this lens. Calculate the position of its image

Options

A. `-10 cm `
B. `+ 10 cm `
C. `-25 cm `
D. `+ 25 cm `

Answer

To solve the problem of finding the position of the image formed by a compound lens with given powers, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Powers of the Lenses**:
- The power of the first lens \( P_1 = 3D \)
- The power of the second lens \( P_2 = -5D \)

2. **Calculate the Equivalent Power of the Compound Lens**:
\[ P_{eq} = P_1 + P_2 = 3D + (-5D) = -2D \]

3. **Find the Focal Length of the Compound Lens**:
The focal length \( F \) is related to the power \( P \) by the formula:
\[ F = \frac{1}{P_{eq}} = \frac{1}{-2D} = -\frac{1}{2} \text{ meters} = -50 \text{ cm} \]

4. **Identify the Object Distance**:
The object distance \( u \) is given as:
\[ u = -50 \text{ cm} \quad (\text{object distance is taken as negative in lens formula}) \]

5. **Apply the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]

6. **Substituting the Values**:
Substitute \( f = -50 \text{ cm} \) and \( u = -50 \text{ cm} \):
\[ \frac{1}{v} = \frac{1}{-50} + \frac{1}{-50} \]
\[ \frac{1}{v} = -\frac{1}{50} - \frac{1}{50} = -\frac{2}{50} = -\frac{1}{25} \]

7. **Calculate the Image Distance \( v \)**:
Taking the reciprocal gives:
\[ v = -25 \text{ cm} \]

### Final Result:
The position of the image is at \( -25 \text{ cm} \), which means the image is formed 25 cm on the same side as the object.