Given
Object distance, u=−15 cm
Focal length, f=+10 cm
Object height, h=+2 cm
Image distance, v=?
Image height, h′=?
Using the lens formula,
1v−1u=1f
1v−1−15=1+10
v=30 cm
Positive sign of v shows that the image is formed at a distance of 30 cm on the right side of the lens. Therefore, the image is real and inverted.
From the formula of magnification,
m=vu=30−15=−2
Again, from the formula of magnification,
m=h′h=vu
h′2=+30−15=−2
h′=−2×2=−4 cm
Negative sign with the magnification and height of the image shows that the image is inverted and real. Thus, a real image of height 4 cm is formed at a distance of 30 cm on the right side of the lens. Image is inverted and twice the size of the object.