Question

Part A: A plastic lens is made with material with index of refraction n=1.4. If the front lens is ground to a radius of 1.3 m and the rear lens ground to a radius of -0.9 m, what is the refractive power of the lens? Answer: 0.75 diopters

Part B: A plastic lens is made with material with index of refraction n=1.4. If the front lens is ground to a radius of 1.3 m and the rear lens ground to a radius of -0.9 m, what is the focal length of the lens? Answer: 1.5m

Part C: A 1.2 cm figurine is placed 0.8 m in front of the lens in the previous problem. Where will the image be?

Options:

- (1) 0.58m from the lens on opposite side from object

-(2) 0.58m from lens on same side as object

-(3) 1.7m from lens on opposite side from object

-(4) 1.7m from lens on same side as object

Part D: A 1.2 cm figurine is placed 0.8 m in front of the lens in the previous problem. Will the image be upright or inverted?

Part E: A 1.2 cm figurine is placed 0.8 m in front of the lens in the previous problem. What will the height of the image be? You may take the absolute value of the image height.

Options:

(1) 2.6cm

(2) 1.2cm

(3) 2.1 cm

(4) 8.4cm

Answer 1

part a:

let radius of front side Rf and radius of back side be Rb.

index of refraction be n and focal length be f.

then focal length is related by:

1/f=(n-1)*((1/Rf)-(1/Rb))

given values are:

Rf=1.3 m

Rb=-0.9 m

n=1.4

then 1/f=0.75214

==>f=1.3295 m

refractive power=1/f=0.75214 diopter

part b:

focal length=1.3295 m

part C:

object distance=u=-0.8 m

focal length=f=1.5 m

let image distance v.

using thin lens equation:

(1/v)-(1/u)=1/f

==>v=-1.7143 m

negative sign shows tht the image is on the same side of the object.

hence 4th option is correct.

part D:

magnification=image height/object height=-image distance/object distance

==>image height/1.2=-(-1.7143)/(-0.8)

==>image height=-2.143

negative sign shows the image is inverted.

part E:

as found in part D, image is 2.143 m high

hence third option is correct.