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

cm

18.4 cm

9.2 cm

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

real /virtual    

Is the image for the object at distance 18.4 real or virtual?

real /virtual    

Is the image for the object at distance 9.2 real or virtual?

real / virtual    

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

upright / inverted    

Is the image for the object at distance 18.4 upright or inverted?

upright / inverted    

Is the image for the object at distance 9.2 upright or inverted?

upright / inverted    

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


Find the magnification for the object at distance 18.4 cm.


Find the magnification for the object at distance 9.2 cm.

cm

Explanation / Answer

For diverging lens f = -18.4 cm

Given that object distance is u = 36.8 cm

image distance is v = ?

A) apply lense equation

1/f = (1/u)+(1/v)

-1/18.4 = (1/36.8)+(1/v)

(1/v) = (-1/18.4)-(1/36.8)

v = -12.26 cm

distance is 12.26 cm

infront of the lens

For 18.4 cm

1/v = (-1/f)-(1/u) =(-1/18.4)-(1/18.4)

v = -9.2 cm

distance is 9.2 cm

infront of the lens

for 9.2 cm

1/v = (-1/f)-(1/u) = (-1/18.4)-(1/9.2)

v = -6.13 cm

distance is 6.13 cm

infront of the lens

B) For 36.8 cm

virtual

For 18.4 cm

virtual

For 9.2 cm

virtual

C) For 36.8 cm

Upright

For 18.4 cm

Upright

For 9.2 cm

Upright

d) Magnification m = -v/u

For 36.8 cm

m = -(-12.26)/36.8
m = 0.333

For 18.4 cm

m = -v/u = -(-9.2)/18.4 = 0.5

For 9.2 cm

m =-v/u = 6.13/9.2 = 0.666

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