- #1

- 75

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my answer:

1/f = (n-1)(1/R1-1/R2)

1/f = (1.5-1)(1/20-1/40)

1/f = .5(.025)

f = 1/.0125 = 80cm (convering lens because f is positive)

Mt=-f/x0=-80/40=-2cm (Inverted image because Mt is negative)

Did I use the correct equations?

- Thread starter jlmac2001
- Start date

- #1

- 75

- 0

my answer:

1/f = (n-1)(1/R1-1/R2)

1/f = (1.5-1)(1/20-1/40)

1/f = .5(.025)

f = 1/.0125 = 80cm (convering lens because f is positive)

Mt=-f/x0=-80/40=-2cm (Inverted image because Mt is negative)

Did I use the correct equations?

- #2

- 508

- 0

Looks O.K. to me.Originally posted by jlmac2001

f = 1/.0125 = 80cm

I should expect, if the object is only 40cm from the lens while the lense's focal length is 80cm, then the image should be upright and virtual, and magnified by a factor of 2.Mt=-f/x0=-80/40=-2cm

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