Question -
Answer -
Size of the object, h1 = 3 cm
Object distance, u = −14 cm
Focal length of the concave lens, f =−21 cm
Image distance = v
According to the lensformula, we have the relation:
Hence, the image is formedon the other side of the lens, 8.4 cm away from it. The negative sign showsthat the image is erect and virtual.
Themagnification of the image is given as:
Hence, the height of theimage is 1.8 cm.
Ifthe object is moved further away from the lens, then the virtual image willmove toward the focus of the lens, but not beyond it. The size of the imagewill decrease with the increase in the object distance.