An insect 550 mm tall is placed 200 cm to the left of a thin

An insect 5.50 mm tall is placed 20.0 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 12.2 cm , and the index of refraction of the lens material is 1.70.

Part A

Calculate the location of the image this lens forms of the insect.

230.4

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Part B

Calculate the size of the image.

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Part C

Is the image real or virtual? erect or inverted?

The image is Please Choosereal and erect real and inverted virtual and erect virtual and inverted  .

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Solution

A.

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

When R1 tends to infinity, then R2 = -12.2 cm

1/f = (1.70 - 1)*(1/infinity - 1/(-12.2))

1/f = [0.70*(0 + 1/12.2)]

f = [0.70*(0 + 1/12.2)]^-1 = 17.42 cm

Now

1/u + 1/v = 1/f

u = 20 cm

f = 17.42 cm

v = uf/(u - f)

v = 17.42*20/(20 - 17.42)

v = 135.03 cm

B. m = -v/u = -135.03/20 = -6.75

m = hi/ho

hi = m*ho = -6.75*5.5 mm = -37.125 mm

C. v is positive means image is real

hi is negative means image is inverted

Let me know if you have any doubt.