A converging lens with a focal length of 12.0cm is placed 30.0cm to the left of

Question

A converging lens with a focal length of 12.0cm is placed 30.0cm to the left of a diverging lens with a focal length of -8.00cm. A candle is placed 36.0cm to the left of the converging lens.

a.) Where is the intermediate image of the candle created by the converging lends alone located?

b.) Where is the final image of the two-lens system located relative to the diverging lens?

c.) What is the total magnification of the final image?

d.) Is the final image real or virtual? Upright or inverted? Justify your answers

e.) Draw a ray diagram roughly to scale showing the formation of the final image by the diverging lens. Include the object(the intermediate image), the final image, the lens and all three principal rays. The diagram should be at least half a letter-size page in size to get full credit

Thank you!!

Explanation / Answer

a)

image due to first lense(converging lens)

object distance u=36 cm

f1=12cm

let image distance be v

1/u+1/v=1/f1

1/36+1/v=1/12

===>

v=18cm is the image distance due to first lense

b)

now,

object distance for the second lense(diverging lens) is,

u'=30-18 cm

u'=12 cm

let image distnace be v'

and

f2=-8 cm

now

1/u'+1/v'=1/f2

1/12+1/v'=1/-8

===>

v'=-4.8cm

image distance due to second lense is v'=-4.8 cm

c)

here

total magnification M=m1*m2=h'/h

here,

object height h=1.4cm

final image height is h'

and

magnificatin of first lense is m1=-v/u=-(-18/36)=0.5

and

magnificatin of second lense is m2=-v'/u'=-(-4.82)/12 =0.402

now,

M=m1*m2

=0.5*0.403

=0.201 kg

d)

overal magenification, M=0.201 kg

final image is virual and upright

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