We are given that:
Focal length of the lens \( f = 10 \, \text{cm} \),
Object distance \( u = 15 \, \text{cm} \).
To determine the image characteristics, we use the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where:
\( v \) is the image distance,
\( u \) is the object distance,
\( f \) is the focal length.
Step 1: Substitute the known values into the lens formula: \[ \frac{1}{10} = \frac{1}{v} - \frac{1}{15} \]
Step 2: Solve for \( v \): \[ \frac{1}{v} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \] \[ v = 6 \, \text{cm} \]
Step 3: Since the image distance \( v \) is positive, the image is real. The magnification produced by the lens is also negative, which means the image is inverted. Thus, the correct answer is Real and inverted.
Step 4: The object is placed outside the focal length, which results in the formation of a real, inverted image.