A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point `15 cm` from the lens. If the lens is removed, the point where the rays meet, move `5 cm` closer to the mounting that holds the lens. Find the focal length of the lens.

Explanation

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

### Step 1: Understand the Setup
We have a converging beam of rays that passes through a diverging lens (concave lens). After passing through the lens, the rays intersect at a point 15 cm from the lens. When the lens is removed, the point where the rays meet moves 5 cm closer to the lens.

### Step 2: Define Variables
- Let \( v \) be the image distance from the lens, which is given as \( v = 15 \) cm.
- When the lens is removed, the rays meet at a point \( 15 \, \text{cm} - 5 \, \text{cm} = 10 \, \text{cm} \) from the original position of the lens. This distance is the new image distance \( v' = 10 \) cm.
- Let \( u \) be the object distance from the lens.

### Step 3: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length of the lens (which we need to find),
- \( v \) is the image distance (15 cm),
- \( u \) is the object distance.

### Step 4: Relate the Object Distance
When the lens is removed, the rays converge at a distance of 10 cm. This means that the object distance \( u \) can be found using the new image distance. Since the rays converge at a point closer by 5 cm, we can say:
\[ u = v' = 10 \, \text{cm} \]

### Step 5: Substitute Values into the Lens Formula
Now we can substitute the values into the lens formula:
\[ \frac{1}{f} = \frac{1}{15} - \frac{1}{10} \]

### Step 6: Calculate the Right Side
To perform the subtraction, we need a common denominator:
\[ \frac{1}{15} = \frac{2}{30}, \quad \frac{1}{10} = \frac{3}{30} \]
Thus,
\[ \frac{1}{f} = \frac{2}{30} - \frac{3}{30} = -\frac{1}{30} \]

### Step 7: Solve for Focal Length
Now, we can find \( f \):
\[ f = -30 \, \text{cm} \]

### Final Answer
The focal length of the diverging lens is \( -30 \, \text{cm} \).

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