Step-by-step explanation:
To determine the object and image locations and the nature of the image formed by the converging lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length of the lens
v = image distance from the lens (positive for real images, negative for virtual images)
u = object distance from the lens (positive for objects to the left of the lens, negative for objects to the right of the lens)
Given:
f = 8.10 cm (focal length)
u = ?
v = ?
We can use the magnification formula to relate the heights of the object and the image:
m = h'/h = -v/u
where:
m = magnification
h' = height of the image
h = height of the object
Given:
h' = 1.70 cm (height of the image)
h = 5.60 mm = 0.56 cm (height of the object)
Let's solve for the object distance (u) first:
m = -v/u
0.56/1.70 = -v/u
u = -v(0.56/1.70)
Now, let's use the lens formula to find the image distance (v):
1/f = 1/v - 1/u
1/8.10 = 1/v + 1/(-v(0.56/1.70))
Simplifying the equation:
1/8.10 = 1/v - 1.7/(0.56v)
1/8.10 = (0.56v - 1.7)/(0.56v)
0.56v - 1.7 = 8.10
0.56v = 9.80
v = 9.80/0.56
v ≈ 17.50 cm
Substituting the value of v back into the equation for u:
u = -v(0.56/1.70)
u = -(17.50)(0.56/1.70)
u ≈ -5.76 cm
Therefore, the object is located approximately 5.76 cm to the right of the lens, and the image is located approximately 17.50 cm to the right of the lens.
To determine the nature of the image, we can observe that the image is erect (upright), which indicates that it is virtual.