A converging beam of light forms a sharp image on a screen. A lens is placed in the path of the beam at `10 cm` from the screen. It is found that the screen has to be moved `8 cm` further away from the lens to obtain a sharp image. Find the focal length and nature of the lens.

Explanation

To solve the problem, we will use the lens formula and the information provided in the question.

### Step 1: Understand the situation
Initially, a converging beam of light forms a sharp image on a screen. When a lens is placed 10 cm from the screen, the screen must be moved 8 cm further away to get a sharp image again.

### Step 2: Define the variables
Let:
- \( d_1 = 10 \, \text{cm} \) (distance from the lens to the original screen position)
- \( d_2 = d_1 + 8 = 10 + 8 = 18 \, \text{cm} \) (distance from the lens to the new screen position)
- \( u = -10 \, \text{cm} \) (object distance for the lens, taken as negative as per the sign convention)
- \( v = -18 \, \text{cm} \) (image distance for the lens, also taken as negative)

### Step 3: Apply the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{-18} - \frac{1}{-10} \]

### Step 4: Calculate the right-hand side
Calculating the right-hand side:
\[ \frac{1}{f} = -\frac{1}{18} + \frac{1}{10} \]
Finding a common denominator (which is 90):
\[ \frac{1}{f} = -\frac{5}{90} + \frac{9}{90} = \frac{4}{90} \]

### Step 5: Solve for focal length \( f \)
Now, we can find \( f \):
\[ f = \frac{90}{4} = 22.5 \, \text{cm} \]

### Step 6: Determine the nature of the lens
Since the focal length \( f \) is positive, the lens is a **converging lens** (convex lens).

### Final Answer
The focal length of the lens is \( 22.5 \, \text{cm} \) and the nature of the lens is **converging**.

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