A diverging lens has a focal length of magnitude 18.4 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 18.4 cm.

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

---Select--- in front of behind

(b) Is the image for the object at distance 36.8 real or virtual?

real virtual   

(c) Is the image for the object at distance 36.8 upright or inverted?

uprightinverted    

(d) Find the magnification for the object at distance 36.8 cm.

---Select--- in front of behind

(b) Is the image for the object at distance 36.8 real or virtual?

real virtual   

(c) Is the image for the object at distance 36.8 upright or inverted?

uprightinverted    

(d) Find the magnification for the object at distance 36.8 cm.

Explanation / Answer

apply for diverging lens -1/f = 1/u + 1/v

u = oject distance,

v= image distance

and f = focal length

magnifiication m = -(v/u) =y'/y ) =y'/y

so here u = 36.8 cm

v = ?

f = 18.4 cm

so 1/v = 1/f - 1/u

v = uf/(u-f)

v = (-36.8 * 18.4)/(36.8 + 18.4)

v = -12.26 cm is the image location

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b. as v is -ve, it must be virtual

--------------------------------------

c. v is -ve, so it is inverted

------------------------------------------

d. magnification m = -v/u

m = -12.26/36.8

m = -0.334

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