Question

An insect 3.60 mm tall is placed 22.8 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 12.8 cm, and the index of refraction of the lens material is 1.64.

(a) Calculate the location of the image this lens forms of the insect.

distance	cm
location	---Select--- to the left of the lens to the right of the lens

Calculate the size of the image.
mm

Is it real or virtual?

Erect or inverted?

(b) Repeat part (a) if the lens is reversed.
Calculate the location of the image this lens forms of the insect.

distance	cm
location	---Select--- to the left of the lens to the right of the lens

Calculate the size of the image.
mm

Is it real or virtual?

Erect or inverted?

Answer 1

object distance s = 22.8 cm

height y = 3.6 mm

Radius of curvature of left surface R1 = infinity

Radius of curvature of right surface R2 = -12.8 cm

from thin lens equation

1/s + 1/s' = (n-1)*(1/R1 - 1/R2)

1/22.8 + 1/s' = (1.64-1)*(1/infinity + 1/12.8)

image distacne s' = 163 cm

size of image y' = y*(s'/s) = 3.6*(163/22.8) = 25.7 mm

real

inverted

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(b)
===

object distance s = 22.8 cm

height y = 3.6 mm

Radius of curvature of left surface R1 = 12.8 cm

Radius of curvature of right surface R2 = infinity

from thin lens equation

1/s + 1/s' = (n-1)*(1/R1 - 1/R2)

1/22.8 + 1/s' = (1.64-1)*(1/12.8 - 1/infinity )

image distacne s' = 163 cm

size of image y' = y*(s'/s) = 3.6*(163/22.8) = 25.7 mm

real

inverted

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