Question

Question 4: A 4.00 cm long object is placed 100. cm in front of a diverging lens as shown in figure. The magnitude of the focal length is 50.0 cm. Find the location of the image s', magnification m, and height h'. Trace two light rays starting at the object and forming the real or virtual image.

Answer 1

Given,

Object distance, u = 100 cm

object height, h = 4.00cm

focal length of the lens, f = 50.0cm

We have the len's formula, 1/f = (1/v) – (1/u)

–1/50 = (1/v) – (–1/100)

–1/50 = (1/v) +(1/100)

1/v = –1/100–1/50

1/v = (–1–2)/100

1/v =  –3/100

v = –100/3

Image distance from the lens,v = 33.3cm

So, the image distance from the lens is 33.3cm, and negative sign indicates that the image is formed on the same side as the object.

Magnification m = v/u

m = (–33.3)/(–100)

m = 0.33

Magnification is less than zero, which means the image is diminished(image size is less than that of the object), and the positive sign of m indicates that the image is erect and virtual.

We have the formula for magnification, m = image size/ object size = h′/h

  0.33 = h′/4

Height of the image, h′= 1.32cm

Object Image -f=50145