A diverging lens has a focal length of magnitude 19.2 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 19.2 cm.

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

19.2 cm

9.6 cm

(b) Is the image for the object at distance 38.4 real or virtual?

realvirtual    

Is the image for the object at distance 19.2 real or virtual?

realvirtual    

Is the image for the object at distance 9.6 real or virtual?

realvirtual    

(c) Is the image for the object at distance 38.4 upright or inverted?

uprightinverted    

Is the image for the object at distance 19.2 upright or inverted?

uprightinverted    

Is the image for the object at distance 9.6 upright or inverted?

uprightinverted    

(d) Find the magnification for the object at distance 38.4 cm.


Find the magnification for the object at distance 19.2 cm.


Find the magnification for the object at distance 9.6 cm.

Explanation / Answer

u = object distance
v = image distance
f = focal length
m = magnification

a) 1 / u + 1 / v = 1 / f
1 / 38.4 + 1 / v = 1 / (19.2)   
v = 38.4cm.
Image distance 38.4 cm.

m = v / u
= 38.4 / 38.4
= 1
Magnification 1.

Image is Virtual ,Upright,Reduced.

Similarly for the remaining

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.