Question

A 3.75mm tall object is located 22.5cm to the left of a planoconvex lens, oriented such that the flat surface faces the object. The second surface of the lens has a radius of curvature with magnitude 13cm. The lens' index of refraction is 1.7. (a) find the location and size of the image. Real? Inverted? (b) repeat the calculation if the lens is reversed, with the convex side now facing the object.

Answer 1

height of the object, ho=3.75mm

object distance, so=22.5cm

radius of curvatur of the convex surfeace, R=13cm

refractive index, n=1.7

a)

use,

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

1/f=(1.7-1)*(0-1/(-13))

===> f=18.57 cm

focal length, f=18.57 cm

and

1/f=1/so + 1/si

1/18.57 = 1/22.5 + 1/si

===> si=106.32 cm

image distance, si=106.32 cm

now,

magnification, m=-di/do = hi/ho

m=-106.32/22.5

m=-4.72

and

m=hi/ho

-4.72=hi/3.75

===> hi=-17.7 cm

size of the image, hi=-17.7cm

image is real and inverted

b)

if the lens is reversed

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

1/f=(1.7-1)*(1/13-0)

===> f=18.57 cm

focal length, f=18.57 cm

and

1/f=1/so + 1/si

1/18.57 = 1/22.5 + 1/si

===> si=106.32 cm

image distance, si=106.32 cm

now,

magnification, m=-di/do = hi/ho

m=-106.32/22.5

m=-4.72

and

m=hi/ho

-4.72=hi/3.75

===> hi=-17.7 cm

size of the image, hi=-17.7cm

image is real and inverted