For an object which is 4.0 mm in height located 6.0 cm to the left of a converging lens
(positive f) of focal length +9.0 cm, determine the image location and characterize it. Show
your work here. No Calculator.
Image is located ________________cm to the ______________ of the lens and the image is
Real/virtual, erect/inverted, and is __________mm in height.
To determine the image location and characterize it, we can use the lens equation:
1/f = 1/dₒ + 1/dᵢ
where f is the focal length of the lens, dₒ is the object distance (distance of the object from the lens), and dᵢ is the image distance (distance of the image from the lens).
Given:
f = +9.0 cm (converging lens)
dₒ = -6.0 cm (object distance, negative as it is to the left of the lens)
To find the image distance, we can rearrange the lens equation:
1/dᵢ = 1/f - 1/dₒ
1/dᵢ = 1/9 - 1/(-6)
Now let's simplify this expression:
1/dᵢ = (2/18) + (3/18)
1/dᵢ = 5/18
Taking the reciprocal of both sides:
dᵢ = 18/5 = 3.6 cm
Since the image distance is positive, the image is formed on the opposite side of the lens, which is to the right.
Now let's characterize the image:
Since the object distance is negative and the image distance is positive, the image is real.
The image distance is smaller than the object distance, so the image is also smaller compared to the object. Since the object is 4.0 mm in height, we can use a magnification equation to find the height of the image:
magnification = height of image / height of object = -dᵢ / dₒ
magnification = (-3.6 cm) / (-6.0 cm) = 0.6
The magnification is positive, indicating that the image is upright or erect.
To find the height of the image, we multiply the magnification by the height of the object:
height of image = magnification * height of object
height of image = 0.6 * 4.0 mm = 2.4 mm
Therefore, the image is located 3.6 cm to the right of the lens, is real, erect, and has a height of 2.4 mm.