In case of displacement method of lenses, the product of magnification in both cases is

Options

A. 1
B. 2
C. Zero
D. Infinite

Answer

To solve the problem regarding the product of magnification in the displacement method of lenses, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Magnification**:
- Magnification (M) is defined as the ratio of the height of the image (h') to the height of the object (h). It can also be expressed in terms of the object distance (u) and the image distance (v) as:
\[ M = \frac{h'}{h} = \frac{v}{u} \]

2. **Defining Magnifications for Two Cases**:
- Let \( M_1 \) be the magnification for the first case, which can be expressed as:
\[ M_1 = \frac{v_1}{u_1} \]
- Let \( M_2 \) be the magnification for the second case, which can be expressed as:
\[ M_2 = \frac{u_2}{v_2} \]
- In the displacement method, the object and image distances interchange when the lens is moved.

3. **Relating the Distances**:
- When the lens is displaced, the object distance \( u \) and image distance \( v \) switch roles. Therefore, we can write:
\[ u_1 = v_2 \quad \text{and} \quad v_1 = u_2 \]

4. **Calculating the Product of Magnifications**:
- Now, we can find the product of the two magnifications:
\[ M_1 \times M_2 = \left(\frac{v_1}{u_1}\right) \times \left(\frac{u_2}{v_2}\right) \]
- Substituting \( u_2 \) and \( v_2 \) with \( v_1 \) and \( u_1 \):
\[ M_1 \times M_2 = \left(\frac{v_1}{u_1}\right) \times \left(\frac{v_1}{u_1}\right) \]
- This simplifies to:
\[ M_1 \times M_2 = \frac{v_1 \cdot v_1}{u_1 \cdot u_1} = 1 \]

5. **Conclusion**:
- The product of the magnifications in both cases is:
\[ M_1 \times M_2 = 1 \]
- Therefore, the correct answer is option **1**.