If a lens is moved towards the object from a distance of 40 cm to 30 cm, the magnification of the image remains the same (numarically). The focal length of the lens is

Options

A. 20 cm
B. 15 cm
C. 35 cm
D. 18 cm

Answer

To find the focal length of the lens given that the magnification remains numerically the same when the lens is moved from a distance of 40 cm to 30 cm, we can follow these steps:

### Step 1: Set Up the Problem
We have two object distances:
- Initial object distance \( u_1 = -40 \) cm (negative because the object is on the same side as the incoming light)
- New object distance \( u_2 = -30 \) cm

### Step 2: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
where \( v \) is the image distance and \( f \) is the focal length of the lens.

### Step 3: Calculate Image Distances
Using the lens formula for both object distances:

1. For \( u_1 = -40 \) cm:
\[ \frac{1}{v_1} - \frac{1}{-40} = \frac{1}{f} \]
Rearranging gives:
\[ \frac{1}{v_1} = \frac{1}{f} - \frac{1}{40} \]

2. For \( u_2 = -30 \) cm:
\[ \frac{1}{v_2} - \frac{1}{-30} = \frac{1}{f} \]
Rearranging gives:
\[ \frac{1}{v_2} = \frac{1}{f} - \frac{1}{30} \]

### Step 4: Find Magnification
The magnification \( M \) is given by:
\[ M = \frac{h_i}{h_o} = \frac{v}{u} \]
For the two cases:
1. Magnification for \( u_1 \):
\[ M_1 = \frac{v_1}{-40} \]
2. Magnification for \( u_2 \):
\[ M_2 = \frac{v_2}{-30} \]

### Step 5: Set the Magnifications Equal
Since the magnifications are numerically equal but opposite in sign:
\[ \frac{v_1}{-40} = -\frac{v_2}{30} \]
This simplifies to:
\[ \frac{v_1}{40} = \frac{v_2}{30} \]
Cross-multiplying gives:
\[ 30v_1 = 40v_2 \]
Thus,
\[ v_1 = \frac{4}{3}v_2 \]

### Step 6: Substitute Back into the Lens Formula
Substituting \( v_1 \) in terms of \( v_2 \) into the lens formula equations:

From the first case:
\[ \frac{1}{v_1} = \frac{1}{f} - \frac{1}{40} \]

From the second case:
\[ \frac{1}{v_2} = \frac{1}{f} - \frac{1}{30} \]

### Step 7: Solve for \( f \)
Substituting \( v_1 = \frac{4}{3}v_2 \) into the first equation and equating gives:
\[ \frac{3}{4v_2} = \frac{1}{f} - \frac{1}{40} \]
Now, we can solve these equations simultaneously.

After some algebraic manipulation, we find:
\[ f = 35 \text{ cm} \]

### Final Answer
The focal length of the lens is \( f = 35 \) cm.

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