Question

An insect 3.85 mm tall is placed 22.6 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 13.4 cm, and the index of refraction of the lens material is 1.60.

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

Calculate the size of the image.

(b) Repeat part (a) if the lens is reversed.

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

Calculate the size of the image.

Answer 1

a) given, R1 = inifite

R2 = -13.4 cm

n = 1.6

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

= (1.6-1)*(1/infinite - 1/(-13.4))

1/f = 0.6*(0 +1/13.4)

f = 13.4/0.6

= 22.33 cm

object distance, u = 22.6 cm

let v is the imeg distance.

Apply, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/22.33 - 1/22.6

v = 1869 cm <<<<<---------Answer

magnificationm ,m = -v/u

= -1869/22.6

= -82.7

image size = m*object size

= 82.7*3.85

= 318 mm or 31.8 cm or 0.318 m <<<<<---------Answer

b) when lense is reversed

R1 = 13.4 cm

R2 = infinte

n = 1.6

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

= (1.6-1)*(1/13.4 - 1/infinite)

1/f = 0.6*(0 +1/13.4)

f = 13.4/0.6

= 22.33 cm

object distance, u = 22.6 cm

let v is the imeg distance.

Apply, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/22.33 - 1/22.6

v = 1869 cm <<<<<---------Answer

magnificationm ,m = -v/u

= -1869/22.6

= -82.7

image size = m*object size

= 82.7*3.85

= 318 mm or 31.8 cm or 0.318 m <<<<<---------Answer