Question

A thin converging lens is to focus the image of an object onto a screen so that the image is life-sized. The lens surfaces have equal radii of curvature of 112 cm, and the refractive index of the lens material is 1.500. (a) How far from the lens should the screen be placed? (b) How far is the screen from the object?

i'm confused

in this problem, is the magnification supposed to be +1 or -1?

for part b) you multiply 2 times the distance of screen from lens? please help!

Answer 1

Radius of curvature of each surface, R = 1.12 m

Refractive index, n = 1.5

The focal length of the lens is given by lens maker’s formula,1/f = (n-1) (1/R1 – 1/R2) =(n-1) 2/R

1/f = ( 1.5 – 1 )x 2 /1.12= 1/1.12

Focal length of the converging lens, f = 1.12 m

(a)

If the image is life-sized, the height of the image is same as that of the object. This implies that the linear magnification is m = -1

If u and v are the object and image distance then, v/u = -1 which gives u = - v

Using the lens formula, 1/f = 1/v – 1/u

1/f = 1/v – 1/-v = 2/v

Image distance, v = 2f = 2x 1.12 = 2.24 m

Distance of the screen from the lens is v = 2.24 m

(b)

Distance between the screen and object = u + v = 2.24 + 2.24 = 4.48 m ( the numerical value of u and v are same and is equal to 2f to get a magnification of m = - 1)

  Distance between the screen and object = 4.48 m