Question

A diverging lens has a focal length of -30.0cm. Located the images for object distances of (a) 60.0cm (b) 30.0cm (c) 15.0cm. What kind of images are they (real or virtual; upright or inverted)? (d) Draw a ray diagram for part (a).

Answer 1

for p =60cm,

1/60 + 1/q = -1/30

q = - 1 cm (q isnegative, so image is in front of lens

Maginication M = -q/p = - (-20/60) <---positivevalue of M means image is upright

when q is negative images are virtual

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for p = 30

1/p + 1/q = -1./f

1/30 +1/q = -1/30

q = - 1/0.0667 cm (q isnegative, so image is in front of lens

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for p = 15

1/15 + 1/q = -1/30

q = -1/0.1 (q is negative,so image is in front of lens
  

for p =60cm,

1/60 + 1/q = -1/30

q = - 1 cm (q isnegative, so image is in front of lens

Maginication M = -q/p = - (-20/60) <---positivevalue of M means image is upright

when q is negative images are virtual

-------------------------------------------------------
for p = 30

1/p + 1/q = -1./f

1/30 +1/q = -1/30

q = - 1/0.0667 cm (q isnegative, so image is in front of lens

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

for p = 15

1/15 + 1/q = -1/30

q = -1/0.1 (q is negative,so image is in front of lens