An insect 5.50 mm tall is placed 20.0 cm to the left of a thin planoconvex lens.

Question

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

SubmitMy AnswersGive Up

Incorrect; Try Again; 3 attempts remaining

Part B

Calculate the size of the image.

SubmitMy AnswersGive Up

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  .

SubmitMy AnswersGive Up

230.4

Explanation / Answer

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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.