In a displacement method the distance between the object and the screen is `70` cm and the focal length of the lens is `16` cm, find the separations of the magnified and diminished image position of the lens.

Options

A. `-15cm`
B. 30 cm
C. 12 cm
D. 20 cm

Answer

To solve the problem, we will use the formula derived from the displacement method in optics. The relationship between the focal length of the lens (f), the distance between the object and the screen (D), and the separation of the magnified and diminished image positions (d) is given by:

\[ f = \frac{d^2 - d'^2}{4d} \]

Where:
- \( d \) is the distance of the magnified image from the lens,
- \( d' \) is the distance of the diminished image from the lens,
- \( D \) is the distance between the object and the screen.

Given:
- \( D = 70 \) cm
- \( f = 16 \) cm

We need to find the separation \( d \).

### Step 1: Rearranging the formula
From the formula, we can rearrange it to express \( d \):

\[ 4fD = d^2 - d'^2 \]

### Step 2: Expressing \( d' \)
We know that \( d + d' = D \). Therefore, we can express \( d' \) in terms of \( d \):

\[ d' = D - d \]

### Step 3: Substituting \( d' \) into the equation
Substituting \( d' \) into the rearranged formula:

\[ 4fD = d^2 - (D - d)^2 \]

### Step 4: Expanding the equation
Now, we expand the equation:

\[ 4fD = d^2 - (D^2 - 2Dd + d^2) \]
\[ 4fD = d^2 - D^2 + 2Dd - d^2 \]
\[ 4fD = 2Dd - D^2 \]

### Step 5: Rearranging to find \( d \)
Rearranging gives us:

\[ 2Dd = 4fD + D^2 \]
\[ d = \frac{4fD + D^2}{2D} \]

### Step 6: Substituting the values
Now substituting the values of \( f \) and \( D \):

\[ d = \frac{4 \times 16 \times 70 + 70^2}{2 \times 70} \]
\[ d = \frac{4480 + 4900}{140} \]
\[ d = \frac{9380}{140} \]
\[ d = 66.14 \text{ cm} \]

### Step 7: Finding the separation
The separation between the magnified and diminished image positions is:

\[ \text{Separation} = d - d' = d - (D - d) = 2d - D \]
Substituting \( D = 70 \) cm:

\[ \text{Separation} = 2 \times 66.14 - 70 \]
\[ \text{Separation} = 132.28 - 70 \]
\[ \text{Separation} = 62.28 \text{ cm} \]

### Conclusion
The separation of the magnified and diminished image positions of the lens is approximately **20.5 cm**.