A parallel beam of rays is incident on a consisting pf three thin lenses with a common optical axis. The focal length of the lenses are equal to `f_1=+10 cm and f_2=-20 cm, and f_3=+9 cm` respectively. The distance between the first and the second lens is `15 cm` and between the second and the third is `5 cm.` Find the position of the point at which the beam converges when it leaves the system of lenses.

Explanation

To find the position of the point at which a parallel beam of rays converges after passing through a system of three thin lenses, we will follow these steps:

### Step 1: Analyze the first lens
- **Given**: Focal length of the first lens \( f_1 = +10 \, \text{cm} \).
- Since the incoming rays are parallel, the object distance \( u_1 \) for the first lens is taken as \( -\infty \) (we consider it negative as per the sign convention).
- **Using the lens formula**:
\[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \]
Substituting the values:
\[ \frac{1}{v_1} - 0 = \frac{1}{10} \]
Thus, \( v_1 = 10 \, \text{cm} \). This means the image formed by the first lens is at \( 10 \, \text{cm} \) to the right of the first lens.

### Step 2: Determine the object distance for the second lens
- The distance between the first and second lenses is \( 15 \, \text{cm} \).
- The image formed by the first lens is \( 10 \, \text{cm} \) from it, so the object distance \( u_2 \) for the second lens is:
\[ u_2 = 15 - 10 = 5 \, \text{cm} \quad (\text{to the left of the second lens, hence } u_2 = -5 \, \text{cm}) \]
- **Given**: Focal length of the second lens \( f_2 = -20 \, \text{cm} \).
- **Using the lens formula**:
\[ \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \]
Substituting the values:
\[ \frac{1}{v_2} - \left(-\frac{1}{5}\right) = -\frac{1}{20} \]
Rearranging gives:
\[ \frac{1}{v_2} + \frac{1}{5} = -\frac{1}{20} \]
\[ \frac{1}{v_2} = -\frac{1}{20} - \frac{4}{20} = -\frac{5}{20} = -\frac{1}{4} \]
Thus, \( v_2 = -4 \, \text{cm} \). This means the image formed by the second lens is \( 4 \, \text{cm} \) to the left of the second lens.

### Step 3: Determine the object distance for the third lens
- The distance between the second and third lenses is \( 5 \, \text{cm} \).
- The image formed by the second lens is \( 4 \, \text{cm} \) to the left of the second lens, hence the object distance \( u_3 \) for the third lens is:
\[ u_3 = 5 + 4 = 9 \, \text{cm} \quad (\text{to the left of the third lens, hence } u_3 = -9 \, \text{cm}) \]
- **Given**: Focal length of the third lens \( f_3 = +9 \, \text{cm} \).
- **Using the lens formula**:
\[ \frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3} \]
Substituting the values:
\[ \frac{1}{v_3} - \left(-\frac{1}{9}\right) = \frac{1}{9} \]
Rearranging gives:
\[ \frac{1}{v_3} + \frac{1}{9} = \frac{1}{9} \]
\[ \frac{1}{v_3} = 0 \]
Thus, \( v_3 = \infty \). This means the rays are parallel after passing through the third lens.

### Conclusion
The final image formed by the system of lenses is at infinity, indicating that the parallel rays do not converge at any point but continue to travel parallel.

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