A converging lens has a focal length of `30 cm.` Rays from a `2.0 cm` high filament that pass through the lens from a virtual image at a distance of `50 cm` from the lens. Where is the filament located? What is the height of the image?

Explanation

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

### Step 1: Identify the given values
- Focal length of the lens (f) = +30 cm (positive for a converging lens)
- Distance of the virtual image (v) = -50 cm (negative because the image is virtual)
- Height of the object (h_o) = 2.0 cm

### Step 2: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where:
- \( f \) = focal length
- \( v \) = image distance
- \( u \) = object distance (which we need to find)

Substituting the known values into the lens formula:
\[ \frac{1}{30} = \frac{1}{-50} + \frac{1}{u} \]

### Step 3: Rearranging the equation
Rearranging the equation to solve for \( \frac{1}{u} \):
\[ \frac{1}{u} = \frac{1}{30} + \frac{1}{50} \]

### Step 4: Finding a common denominator
The common denominator for 30 and 50 is 150. Therefore:
\[ \frac{1}{30} = \frac{5}{150}, \quad \frac{1}{50} = \frac{3}{150} \]
So,
\[ \frac{1}{u} = \frac{5}{150} + \frac{3}{150} = \frac{8}{150} \]

### Step 5: Solve for \( u \)
Now, we can find \( u \):
\[ u = \frac{150}{8} = 18.75 \text{ cm} \]
Since the object is on the same side as the incoming light, we take it as negative:
\[ u = -18.75 \text{ cm} \]

### Step 6: Calculate the magnification
The magnification (m) is given by:
\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
Where:
- \( h_i \) = height of the image
- \( h_o \) = height of the object

Substituting the known values:
\[ m = \frac{-50}{-18.75} = \frac{50}{18.75} = \frac{8}{3} \]

### Step 7: Find the height of the image
Now we can find the height of the image:
\[ h_i = m \cdot h_o = \frac{8}{3} \cdot 2 = \frac{16}{3} \text{ cm} \approx 5.33 \text{ cm} \]

### Final Answers
1. The object (filament) is located at approximately **-18.75 cm** from the lens.
2. The height of the image is approximately **5.33 cm**.

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