A converging lens forms a five fold magnified image of an object. The screen is moved towards the object by a distance `d=0.5 m,` and the lens is shifted so that the image has the same size as the object. Find the power of lens and the initial distance between the object and the screen.

Explanation

To solve the problem step by step, we will break it down into manageable parts.

### Step 1: Understand the Magnification
The magnification (M) of a lens is given by the formula:
\[ M = \frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance. According to the problem, the lens forms a five-fold magnified image:
\[ M = 5 \]
Thus, we can write:
\[ v = 5u \]

### Step 2: Analyze the Shift of the Screen
When the screen is moved towards the object by a distance \( d = 0.5 \, m \), the new image distance becomes equal to the object distance:
\[ v' = u \]
The new image distance \( v' \) can be expressed as:
\[ v' = v - d = 5u - 0.5 \]
Setting this equal to \( u \):
\[ 5u - 0.5 = u \]

### Step 3: Solve for Object Distance \( u \)
Rearranging the equation gives:
\[ 5u - u = 0.5 \]
\[ 4u = 0.5 \]
\[ u = \frac{0.5}{4} = 0.125 \, m \]

### Step 4: Calculate Image Distance \( v \)
Using the value of \( u \) to find \( v \):
\[ v = 5u = 5 \times 0.125 = 0.625 \, m \]

### Step 5: Find the Initial Distance Between Object and Screen
The total initial distance \( D \) between the object and the screen is:
\[ D = u + v = 0.125 + 0.625 = 0.75 \, m \]

### Step 6: Use Lens Formula to Find Focal Length \( f \)
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{0.625} - \frac{1}{0.125} \]
Calculating:
\[ \frac{1}{f} = 1.6 - 8 = -6.4 \]
So,
\[ f = -\frac{1}{6.4} \approx 0.15625 \, m \]

### Step 7: Calculate the Power of the Lens
The power \( P \) of a lens is given by:
\[ P = \frac{1}{f} \]
Where \( f \) is in meters. Thus:
\[ P = \frac{1}{0.15625} \approx 6.4 \, D \]

### Final Results
- The initial distance between the object and the screen is \( 0.75 \, m \).
- The power of the lens is approximately \( 6.4 \, D \).