Question

A diverging lens has a focal length of magnitude 23.6 cm.

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

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

23.6 cm

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

11.8 cm

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

(b) Is the image for the object at distance 47.2 real or virtual?

realvirtual     

Is the image for the object at distance 23.6 real or virtual?

realvirtual     

Is the image for the object at distance 11.8 real or virtual?

realvirtual     

(c) Is the image for the object at distance 47.2 upright or inverted?

uprightinverted     

Is the image for the object at distance 23.6 upright or inverted?

uprightinverted     

Is the image for the object at distance 11.8 upright or inverted?

uprightinverted     

(d) Find the magnification for the object at distance 47.2 cm.


Find the magnification for the object at distance 23.6 cm.


Find the magnification for the object at distance 11.8 cm.

Answer 1

Focal length of lens ,f=-23.6 cm.when the object distance do 47.2cm.from lens formula,1/f=1/do+1/di

Hence,di=f do/(do-f)=-23.6×47 .2/(47.2+23.6)=-15.7333 cm,behind

Similarly if do=23.6cm

di=-23.6×23.6/(23.6+23.6) =-11.8 cm,behind

if do=11.8 cm

di=-23.6×11.8/(11.8+23.6)=-7.8666 cm

b) In the three cases images are virtual as the image distance is negative.

C)They are upright

d)magnification,M=-(di/do)

M1=-(-15.7333/47.2)=0.3333326 cm

M2=-(-11.8/23.6)=0.5 cm

M2=-(-7.86666/11.8)=0.66666 cm