A 10-meter object is placed at a distance of 165 meters in front of a lens whose focal length is 50 meters. Which of the following describes the image formed?
To determine the characteristics of the image formed by the 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,
- \(u\) is the object distance.
Given:
- \(f = 50\) meters,
- \(u = 165\) meters.
We can substitute these values into the lens formula to find \(v\):
\(\frac{1}{50} = \frac{1}{v} - \frac{1}{165}\).
To solve for \(v\), we can use algebraic manipulation:
\(\frac{1}{v} = \frac{1}{50} + \frac{1}{165}\).
By finding the least common denominator and combining the fractions on the right-hand side:
\(\frac{1}{v} = \frac{33}{1650} + \frac{10}{1650}\).
Simplifying:
\(\frac{1}{v} = \frac{43}{1650}\).
Taking the reciprocal of both sides:
\(v = \frac{1650}{43}\).
Calculating this value:
\(v \approx 38.372\) meters.
Now that we have the image distance, we can analyze the image formed. The characteristics of the image depend on the relative positions of the object and the lens. Since the object is located beyond the focal length of the lens (165 meters > 50 meters), the image formed will be a real and inverted image. It will also be smaller in size compared to the object.
To understand the type of image formed by the 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 distance of the image from the lens
- \(u\) is the distance of the object from the lens
In this case, the object distance \(u\) is -165 meters (negative sign indicates that the object is placed on the same side as the incident light).
Plugging in the values, we get:
\(\frac{1}{50} = \frac{1}{v} - \frac{1}{165}\)
To solve for \(v\), we'll rearrange the equation:
\(\frac{1}{v} = \frac{1}{50} + \frac{1}{165}\)
Adding the fractions:
\(\frac{1}{v} = \frac{33+10}{1650}\)
\(\frac{1}{v} = \frac{43}{1650}\)
Multiplying both sides by \(v\):
\(v = \frac{1650}{43}\)
Simplifying:
\(v \approx 38.37\)
Since the value of \(v\) is positive, it means the image is formed on the opposite side of the lens as the object. The image distance is approximately 38.37 meters.
Based on the sign convention for lenses:
- If \(v\) is positive, the image is formed on the opposite side (real image).
- If \(v\) is negative, the image is formed on the same side (virtual image).
Therefore, in this case, since \(v\) is positive, the image formed is a real image.