For a certain lens, the magnification of an object when placed at a distance of 0.15 m is twice of the magnification produced, when the distance was 0.2 m. If in both the situations a real image is formed, then what is the focal length (in cm ) of the lens?

Answer

To solve the problem step by step, we will use the lens formula and the magnification formula.

### Step 1: Understanding the Problem
We are given two object distances (u1 = -0.15 m and u2 = -0.2 m) and the relationship between their magnifications. The magnification for the first distance is twice that of the second distance.

### Step 2: Write Down the Magnification Formula
The magnification (M) produced by a lens is given by:
\[ M = -\frac{v}{u} \]
where \( v \) is the image distance and \( u \) is the object distance.

### Step 3: Set Up the Equations
From the problem, we know:
\[ M_1 = -\frac{v_1}{u_1} \quad \text{and} \quad M_2 = -\frac{v_2}{u_2} \]
Given that \( M_1 = 2M_2 \), we can write:
\[ -\frac{v_1}{u_1} = 2 \left(-\frac{v_2}{u_2}\right) \]

### Step 4: Substitute the Values of u
Substituting \( u_1 = -0.15 \, \text{m} \) and \( u_2 = -0.2 \, \text{m} \):
\[ -\frac{v_1}{-0.15} = 2 \left(-\frac{v_2}{-0.2}\right) \]
This simplifies to:
\[ \frac{v_1}{0.15} = 2 \cdot \frac{v_2}{0.2} \]
\[ \frac{v_1}{0.15} = \frac{2v_2}{0.2} \]
Cross-multiplying gives:
\[ v_1 \cdot 0.2 = 2v_2 \cdot 0.15 \]
\[ 0.2v_1 = 0.3v_2 \]
Thus, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{0.3}{0.2} v_2 = 1.5v_2 \]

### Step 5: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]
We can write two equations using the lens formula for both object distances:
1. For \( u_1 = -0.15 \):
\[ \frac{1}{f} = \frac{1}{-0.15} + \frac{1}{v_1} \]
2. For \( u_2 = -0.2 \):
\[ \frac{1}{f} = \frac{1}{-0.2} + \frac{1}{v_2} \]

### Step 6: Substitute \( v_1 \) in Terms of \( v_2 \)
From the first equation:
\[ \frac{1}{f} = \frac{1}{-0.15} + \frac{1}{1.5v_2} \]
From the second equation:
\[ \frac{1}{f} = \frac{1}{-0.2} + \frac{1}{v_2} \]

### Step 7: Set the Two Equations Equal
Setting the two expressions for \( \frac{1}{f} \) equal to each other:
\[ \frac{1}{-0.15} + \frac{1}{1.5v_2} = \frac{1}{-0.2} + \frac{1}{v_2} \]

### Step 8: Solve for \( v_2 \)
Multiply through by \( -0.15 \cdot -0.2 \cdot 1.5v_2 \) to eliminate the fractions and solve for \( v_2 \).

### Step 9: Calculate the Focal Length
Once \( v_2 \) is found, substitute back into either lens formula to find \( f \).

### Step 10: Convert to Centimeters
Convert the focal length from meters to centimeters by multiplying by 100.

### Final Answer
The focal length \( f \) of the lens is found to be 10 cm.