In the displacement method , a convex lens is placed in between an object and a screen . If one of them magnification is 3 and the displacement of the lens between the two positions is 24 cm , then the focal length of the lens is :-

Options

A. 10 cm
B. 9 cm
C. 6 cm
D. 16/3cm

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Given Information
We are given:
- Magnification (M) = 3
- Displacement of the lens (d) = 24 cm

### Step 2: Relate Magnification to Image and Object Distances
The magnification (M) of a lens is given by the formula:
\[ M = \frac{V}{U} \]
where:
- \( V \) = image distance
- \( U \) = object distance

From the problem, we have:
\[ M = 3 \]
Thus,
\[ \frac{V}{U} = 3 \]
This implies:
\[ V = 3U \]

### Step 3: Relate the Displacement to Object and Image Distances
The displacement of the lens is the difference between the image distance and the object distance:
\[ V - U = 24 \, \text{cm} \]

### Step 4: Substitute the Value of V
Substituting \( V = 3U \) into the displacement equation:
\[ 3U - U = 24 \]
This simplifies to:
\[ 2U = 24 \]
Thus,
\[ U = 12 \, \text{cm} \]

### Step 5: Calculate the Image Distance V
Now, substituting the value of \( U \) back into the equation for \( V \):
\[ V = 3U = 3 \times 12 = 36 \, \text{cm} \]

### Step 6: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{V} + \frac{1}{U} \]
Substituting the values of \( V \) and \( U \):
\[ \frac{1}{f} = \frac{1}{36} + \frac{1}{12} \]

### Step 7: Simplify the Right Side
To add these fractions, we need a common denominator:
\[ \frac{1}{f} = \frac{1}{36} + \frac{3}{36} = \frac{4}{36} \]

### Step 8: Solve for Focal Length f
Now, solving for \( f \):
\[ \frac{1}{f} = \frac{4}{36} \]
Thus,
\[ f = \frac{36}{4} = 9 \, \text{cm} \]

### Final Answer
The focal length of the lens is:
\[ \boxed{9 \, \text{cm}} \]

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