A diverging lens has a focal length of magnitude 24.2 cm. (a) Locate the images

Question

A diverging lens has a focal length of magnitude 24.2 cm. (a) Locate the images for each of the following object distances. 48.4 cm distance cm location? infront or behind 24.2 cm distance cm location? infront or behind 12.1 cm distance cm location? in front or behind

(b) Is the image for the object at distance 48.4 real or virtual? real virtual Is the image for the object at distance 24.2 real or virtual? real virtual Is the image for the object at distance 12.1 real or virtual?

(c) Is the image for the object at distance 48.4 upright or inverted? upright inverted Is the image for the object at distance 24.2 upright or inverted? Is the image for the object at distance 12.1 upright or inverted? upright inverted

(d) Find the magnification for the object at distance 48.4 cm. Find the magnification for the object at distance 24.2 cm. Find the magnification for the object at distance 12.1 cm.

Explanation / Answer

focal length is f = -24.2cm, negative because the lens is diverging.

(a)

we will use the lens maker's formula;

1/v - 1/u = 1/f, where v is image distance, u is object distance and f is focal length.

or, 1/v = 1/u + 1/f, but for a diverign lens f is negative and u is always negative hence, v will always be negative which means image will always be formed behind the lens irrespective of the opbject distance.

So, for all the three object distances, the image will be behind the lens.

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(b)

A diverging lens always produces a virtual image no matter what the object the distance may be.

So for all the three object distances given, the image will be virtual.

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(c)

for a lens h2/h1 = v/u, where h2 is the image height and h1 is the object height

for a divergin lens v and u have same sign, i.e., negative. So h2 and h1 will also have same sign.

So image will always be upright for a diverging lens.

So for all the three object distances given, the image will be upright.

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(d)

magnification m = v/u

u = -48.4 cm, using the lens maker's formula we have

v = uf/(u+f) = (-48.4 cm)(-24.2cm)/(-48.4 cm - 24.2cm)

so, v/u = f/(u + f)

Thus magnification is m = f/(u + f)

for u = -48.4 cm

m = (-24.2cm)/(-48.4 cm - 24.2cm)

or m = 0.33

for u = -24.2 cm

m = (-24.2cm)/(-24.2 cm - 24.2cm)

or, m = 0.5

for u = -12.1 cm

m = (-24.2cm)/(-12.1 cm - 24.2cm)

or, m = 0.66

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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