An insect of height 1.5 cm is 13cm from a converging lens of focal length 13.5 cm. Determine the position, height and properties of the image formed.
To determine the position, height, and properties of the image formed by a converging lens, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance (position)
u = object distance (distance of the insect from the lens)
Given:
Height of the insect (object) = 1.5 cm
Focal length of the lens (f) = 13.5 cm
Distance of the insect from the lens (u) = 13 cm
Let's calculate the position, height, and properties of the image:
Step 1: Calculate the object distance (u):
u = 13 cm
Step 2: Plug the given values into the lens formula to calculate the image distance (v):
1/13.5 = 1/v - 1/13
1/v = 1/13.5 + 1/13
1/v = (13 + 13.5) / (13 * 13.5)
1/v = 26.5 / 175.5
v ≈ 6.63 cm
The image distance (v) is approximately 6.63 cm.
Step 3: Determine the magnification (m):
The magnification is given by the formula: m = v/u
m = 6.63 cm / 13 cm
m ≈ 0.51
The magnification (m) is approximately 0.51.
Step 4: Determine the height of the image (h):
The height of the image (h) is given by the formula: h = m * height of the object
h = 0.51 * 1.5 cm
h ≈ 0.77 cm
The height of the image (h) is approximately 0.77 cm.
Step 5: Determine the orientation and nature of the image:
Since the image distance (v) is positive, the image is formed on the opposite side of the lens as the object, which indicates a real image. The positive magnification value (m) indicates that the image is inverted.
Therefore, the image is formed at a distance of approximately 6.63 cm from the lens, it has a height of approximately 0.77 cm, and it is inverted.