A slide with an image 4 cm `xx` 2 cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. The size of the image on the screen is

Options

A. `40 cm xx 20 cm `
B. `36 cm xx 18 cm `
C. `20 cm xx 40 cm `
D. `10 cm xx 50 cm`

Answer

To solve the problem, we need to find the size of the image formed on the screen by the converging lens. Here are the steps to find the solution:

### Step 1: Understand the given information
- The size of the object (slide) is 4 cm x 2 cm.
- The distance of the object from the lens (u) is -10 cm (the negative sign indicates that the object is placed on the same side as the incoming light).
- The distance of the image from the lens (v) is 1.1 m (or 110 cm) from the slide, which means the total distance from the lens to the image is \( v = 110 + 10 = 120 \) cm.

### Step 2: Use the lens formula
The lens formula relates the object distance (u), image distance (v), and the focal length (f) of the lens:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
- \( v = 120 \) cm
- \( u = -10 \) cm

\[ \frac{1}{f} = \frac{1}{120} - \frac{1}{-10} \]
\[ \frac{1}{f} = \frac{1}{120} + \frac{1}{10} \]
To add these fractions, find a common denominator:
\[ \frac{1}{f} = \frac{1}{120} + \frac{12}{120} = \frac{13}{120} \]
Thus, the focal length \( f \) is:
\[ f = \frac{120}{13} \approx 9.23 \text{ cm} \]

### Step 3: Calculate the magnification
The magnification (m) of the lens is given by the formula:
\[ m = \frac{h'}{h} = -\frac{v}{u} \]
Where:
- \( h' \) is the height of the image,
- \( h \) is the height of the object.

Substituting the values:
\[ m = -\frac{120}{-10} = 12 \]

### Step 4: Find the size of the image
Now, we can find the height of the image \( h' \):
\[ h' = m \cdot h \]
Given that the height of the object \( h = 2 \) cm:
\[ h' = 12 \cdot 2 = 24 \text{ cm} \]

### Final Answer
The size of the image on the screen is **24 cm**.