Question

6. An object is placed 200.0 cm to the left of a diverging lengs (f=-300.0cm). A converging lens (f=+1.0cm) is placed 10.0cm to the right of the diverging lens.

a) Determine the position and magnification of the final image formed by the two-lens system.

b) Does this situation more closely represent the correction necessary for nearsightedness or farsightedness? Explain.

Answer 1

focal length of the first lens is, f1=-300cm

focal length of the second lens is, f2=1cm

separation between two lens is, d=10cm

object distance, u1=200cm

apply lens equation for the first lens

1/u1 + 1/v1=1/f1

1/200+1/v1 =1/-300

=====> v1=-120 cm

image distnace from first lense, v1=-120cm

apply lens equation for the second lens

1/u2 + 1/v2=1/f2

1/d-v2 + 1/v2 = 1/f2

1/(10-(-120))+1/v2=1/1

===> v2=1.001 cm

final image distance from the second lens, v2=1.001cm

and

magnefication,

M=m1*m2

M=(-v1/u1)*(-v2/u2)

M=(-120/200)*(-1.001/(10-(-120))=

M=0.0046

M is +ve

====> final image will be virtual,erected(upright) smaller than object

b)

the above situation is closely represent the correcton of the nearsightdness

becasuse, in correction of nearsightness the diverging lens can be used to from the virtual finally