The focal length of the diverging lens is -30 cm and the height of the image is 1.6 cm.
How can focal length of the diverging lens?
We need to use the thin lens equation with proper sign conventions. For a diverging lens:
Focal Length (f) is negative.
Object Distance (d o) is positive (object is in front of the lens).
Image Distance (d i) is negative (image is virtual and on the same side as the object).
Thin Lens Equation: 1/f = 1/d o + 1/d_i
1/f = 1/15 cm + 1/(-6) cm Solving for f: f = -30 cm (negative sign confirms diverging lens)
Image Height
Magnification: Magnification (m) tells us how much the image is magnified or shrunk compared to the object. m = -d i / d o
m = -(-6 cm) / 15 cm m = 0.4 (positive sign indicates upright image)
Image Height: Image Height = Object Height * Magnification Image Height = 4 cm * 0.4 Image Height = 1.6 cm
Therefore, the focal length of the diverging lens is -30 cm and the height of the image is 1.6 cm.