Question
An illuminated object is placed at a distance of 7vm from a converging lens of focal length 30cm. Determine the image distance
Answers
Answered by
GPT-4o mini
To determine the image distance for an illuminated object placed in front of a converging lens (convex lens), we can use the lens formula:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
Where:
- \(f\) is the focal length of the lens,
- \(v\) is the image distance from the lens,
- \(u\) is the object distance from the lens.
1. **Focal Length**:
The focal length given is \(f = 30 \, \text{cm}\) (positive for a converging lens).
2. **Object Distance**:
The object distance is given as \(u = -7 \, \text{vm} = -700 \, \text{cm}\) (negative as per the sign convention that measures distances against the direction of the incoming light).
3. **Substituting values into the lens formula**:
Now we can substitute the values into the lens formula:
\[
\frac{1}{30} = \frac{1}{v} - \frac{1}{-700}
\]
This simplifies to:
\[
\frac{1}{30} = \frac{1}{v} + \frac{1}{700}
\]
Rearranging gives:
\[
\frac{1}{v} = \frac{1}{30} - \frac{1}{700}
\]
4. **Finding a common denominator**:
The least common multiple of 30 and 700 is 2100. We can rewrite the fractions:
\[
\frac{1}{30} = \frac{70}{2100} \quad \text{and} \quad \frac{1}{700} = \frac{3}{2100}
\]
Thus:
\[
\frac{1}{v} = \frac{70}{2100} - \frac{3}{2100} = \frac{67}{2100}
\]
5. **Calculating \(v\)**:
Inverting gives:
\[
v = \frac{2100}{67} \approx 31.34 \, \text{cm}
\]
Thus, the image distance \(v\) from the lens is approximately **31.34 cm**.
The positive value indicates that the image is formed on the opposite side of the lens from the object, which is typical for a converging lens with a real object.
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
Where:
- \(f\) is the focal length of the lens,
- \(v\) is the image distance from the lens,
- \(u\) is the object distance from the lens.
1. **Focal Length**:
The focal length given is \(f = 30 \, \text{cm}\) (positive for a converging lens).
2. **Object Distance**:
The object distance is given as \(u = -7 \, \text{vm} = -700 \, \text{cm}\) (negative as per the sign convention that measures distances against the direction of the incoming light).
3. **Substituting values into the lens formula**:
Now we can substitute the values into the lens formula:
\[
\frac{1}{30} = \frac{1}{v} - \frac{1}{-700}
\]
This simplifies to:
\[
\frac{1}{30} = \frac{1}{v} + \frac{1}{700}
\]
Rearranging gives:
\[
\frac{1}{v} = \frac{1}{30} - \frac{1}{700}
\]
4. **Finding a common denominator**:
The least common multiple of 30 and 700 is 2100. We can rewrite the fractions:
\[
\frac{1}{30} = \frac{70}{2100} \quad \text{and} \quad \frac{1}{700} = \frac{3}{2100}
\]
Thus:
\[
\frac{1}{v} = \frac{70}{2100} - \frac{3}{2100} = \frac{67}{2100}
\]
5. **Calculating \(v\)**:
Inverting gives:
\[
v = \frac{2100}{67} \approx 31.34 \, \text{cm}
\]
Thus, the image distance \(v\) from the lens is approximately **31.34 cm**.
The positive value indicates that the image is formed on the opposite side of the lens from the object, which is typical for a converging lens with a real object.