A convex lens makes a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object, we again get a real image on the screen which is 16 cm tall. The length of the object must be

Options

A. `1//4 cm`
B. `8 cm`
C. `12 cm`
D. `20 cm`

Answer

To solve the problem, we will use the concept of magnification in lens optics. The magnification (M) of a lens is given by the formula:

\[ M = \frac{H_i}{H_o} = -\frac{V}{U} \]

where:
- \( H_i \) is the height of the image,
- \( H_o \) is the height of the object,
- \( V \) is the image distance,
- \( U \) is the object distance.

### Step-by-step Solution:

1. **Identify the given information:**
- For the first image:
- Height of the image \( H_{i1} = 4 \) cm
- For the second image:
- Height of the image \( H_{i2} = 16 \) cm

2. **Set up the magnification equations:**
- For the first image:
\[ M_1 = \frac{H_{i1}}{H_o} = \frac{4}{H_o} \]
- For the second image:
\[ M_2 = \frac{H_{i2}}{H_o} = \frac{16}{H_o} \]

3. **Relate the two magnifications:**
- Since the lens is shifted but the object remains in the same position, the two magnifications are related by the fact that they are inversely proportional:
\[ M_1 \cdot M_2 = -1 \]
- Substituting the magnifications:
\[ \left(\frac{4}{H_o}\right) \cdot \left(\frac{16}{H_o}\right) = -1 \]

4. **Simplify the equation:**
\[ \frac{64}{H_o^2} = -1 \]
- Since height cannot be negative, we can ignore the negative sign:
\[ 64 = H_o^2 \]

5. **Solve for the height of the object:**
\[ H_o = \sqrt{64} = 8 \text{ cm} \]

### Final Answer:
The length of the object must be **8 cm**.