Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5
1 answer:
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Answer:
(A) therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted
(B) therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted
Explanation:
height of the insect (h) = 5.25 mm = 0.525 cm
distance of the insect (s) = 25 cm
radius of curvature of the flat left surface (R1) = ∞
radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)
index of refraction (n) = 1.7
(A) we can find the location of the image by applying the formula below
where
- s' = distance of the image
- f = focal length
- but we first need to find the focal length before we can apply this formula
f =
f = 17.9 cm
now that we have the focal length we can apply
=\frac{1}{s'}
=\frac{1}{s'}
s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens
magnification = where y' is the height of the image, therefore
y' = x 0.525 = -13.22 cm
therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted
(B) if the lens is reversed, the radius of curvatures would be interchanged
radius of curvature of the flat left surface (R1) = ∞
radius of curvature of the right surface (R2) = 12.5 cm
we can find the location of the image by applying the formula below
where
- s' = distance of the image
- f = focal length
- but we first need to find the focal length before we can apply this formula
f =
f = 17.9 cm
now that we have the focal length we can apply
=\frac{1}{s'}
=\frac{1}{s'}
s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens
magnification = where y' is the height of the image, therefore
y' = x 0.525 = -13.22 cm
therefore the image is
- 63 cm to the right of the lens
- the image size is -13.22 cm
- it is real
- it is inverted