Question

A diverging lens has a focal length of magnitude 24.4 cm.

(a) Locate the images for each of the following object distances.
48.8 cm

distance	cm
location	---Select--- in front of the lens behind the lens

24.4 cm

distance	cm
location	---Select--- in front of the lens behind the lens

12.2 cm

distance	cm
location	---Select--- in front of the lens behind the lens

(b) Is the image for the object at distance 48.8 real or virtual?

realvirtual    

Is the image for the object at distance 24.4 real or virtual?

realvirtual    

Is the image for the object at distance 12.2 real or virtual?

realvirtual    

(c) Is the image for the object at distance 48.8 upright or inverted?

uprightinverted    

Is the image for the object at distance 24.4 upright or inverted?

uprightinverted    

Is the image for the object at distance 12.2 upright or inverted?

uprightinverted    

(d) Find the magnification for the object at distance 48.8 cm.


Find the magnification for the object at distance 24.4 cm.


Find the magnification for the object at distance 12.2 cm.

Answer 1

image distance = v ,

object distance = u ,

focal lenght= f ,

magnification = m = v/u

a)

1/v-1/u= 1/ f

1/v- 1/-48.8 =1/-24.4

u = -16.2667 cm same side of the object virtual , small , upright

magnification   = m= v/u = -16.2667/-48.8 =0.333334016

b) u = -24.4 cm

apply same formula so we got  

u = -12.2 cm  same side of the object virtual , small , upright ​​

magnification =m = 12.2/24.4 = 0.5

c) u =-12.2 cm

apply same formula , so we got  

u = -8.13333 cm  same side of the object virtual , small , upright ​​

magnification   = m= v/u = -8.13333/-12.2 =0.666666393​

let me know in a comment if there is any problem or doubts