Final answer:
The image distance for an ant viewed through a -6.90D lens held 11.5 cm away is -6.415 cm, indicating the image is virtual and on the same side as the ant.
Step-by-step explanation:
The question involves a lens with a power of -6.90D, which means it is diverging. The object distance (do) is the distance from the lens to the ant, which in this case is 11.5 cm. To find the image distance (di), we use the lens formula, given by 1/f = 1/do + 1/di, where f is the focal length.
Since we know the power of the lens (P) is -6.90D, we can convert this to focal length (f) using f = 1/P in meters, which yields f = -0.145 meters or -14.5 cm (negative sign indicates a diverging lens).
Now applying the lens formula:
1/f = 1/do + 1/di
1/(-14.5 cm) = 1/(11.5 cm) + 1/di
Solving for di, we find:
1/di = 1/(-14.5 cm) - 1/(11.5 cm)
1/di = -0.06897 cm-1 - 0.08696 cm-1
1/di = -0.15593 cm-1
di = -6.415 cm
The negative sign indicates that the image is on the same side of the lens as the ant, meaning the image is virtual. Hence, the image distance for the ant viewed through the -6.90D lens held 11.5 cm away is -6.415 cm.