A convex lens forms a real image 16 mm long on a screen. If the lens is shifted to a new position without disturbing the object or the screen then again a real image of length 81 mm is formed. The length of the object must be

Options

A. 48.5 mm
B. 36 mm
C. 6 mm
D. 72 mm

Answer

To solve the problem, we need to use the concept of magnification in optics. The magnification (m) produced by a lens is given by the formula:

\[ m = \frac{\text{Height of Image}}{\text{Height of Object}} = \frac{h_i}{h_o} \]

Where:
- \( h_i \) = height of the image
- \( h_o \) = height of the object

Given:
- The first image height \( h_{i1} = 16 \, \text{mm} \)
- The second image height \( h_{i2} = 81 \, \text{mm} \)

Let the height of the object be \( h_o \).

### Step 1: Write the magnification equations for both cases
For the first case:
\[ m_1 = \frac{h_{i1}}{h_o} = \frac{16}{h_o} \]

For the second case:
\[ m_2 = \frac{h_{i2}}{h_o} = \frac{81}{h_o} \]

### Step 2: Use the displacement law of lenses
According to the displacement law of lenses, the ratio of the magnifications when the lens is shifted is equal to the ratio of the heights of the images formed:

\[ \frac{m_1}{m_2} = \frac{h_{i1}}{h_{i2}} \]

Substituting the values we have:
\[ \frac{m_1}{m_2} = \frac{16/h_o}{81/h_o} = \frac{16}{81} \]

### Step 3: Set up the equation
From the above relationship, we can equate the magnifications:
\[ \frac{16}{81} = \frac{h_{i1}}{h_{i2}} \]

### Step 4: Cross-multiply to solve for \( h_o \)
Since we have already established that:
\[ m_1 = \frac{16}{h_o} \]
\[ m_2 = \frac{81}{h_o} \]

We can set up the equation:
\[ \frac{16}{h_o} \cdot \frac{h_o}{81} = 1 \]

This simplifies to:
\[ \frac{16}{81} = \frac{h_{i1}}{h_{i2}} \]

### Step 5: Solve for \( h_o \)
Now, we can set up the equation:
\[ m_1 \cdot m_2 = \frac{16}{h_o} \cdot \frac{81}{h_o} = 1 \]

Thus:
\[ \frac{16 \cdot 81}{h_o^2} = 1 \]

This leads to:
\[ h_o^2 = 16 \cdot 81 \]

Calculating \( 16 \cdot 81 \):
\[ 16 \cdot 81 = 1296 \]

Taking the square root:
\[ h_o = \sqrt{1296} = 36 \, \text{mm} \]

### Conclusion
The length of the object must be \( 36 \, \text{mm} \).

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