Question

1. You have a converging lens with a focal length of 1.5 cm and a diverging lens with a focal length of 2.5 cm. You also have a binder clip with a height of 1.6 cm.

A. Sketch the ray tracing if the binder clip is 4.1 cm from the converging lens.

B. Use the thin lens equation to calculate the location and size of the image if the binder clip is 4.1 cm from the converging lens.

C. What is the refractive power of each lens?

D. What is the total refractive power and focal length if the 2 lenses are used together?

E. If the binder clip is placed 4.1 cm in front of the two lenses together where is the image? Is it real or virtual? Upright or inverted?

Answer 1

The following diagram is the ray diagram of the binder clip with converging lens The image formed by the diverging lens is shown below, RA Calculate the image distance by using the thin lens formula, which is given as, 1 1 1 Substitute -4.1 cm for u and 1.5 cm for fin above equation as follows, 1 1 1 1 1 1 (1.5 cm)(-4.1 cm) 1.5 cm+(-4.1 cm) -2.4 cm

c) Refractive power is given by the following expression as follows, D=1 (1.5 cm) 100 cm) r 1m = +67 m- Refractive power of diverging lens is given as follows, D=-2.5cm D- -2.5 cm 1 m (-2.5 cm) 100 cm = -40 m D) Total refractive power of the combined lens is given as follows, D= +67 m - 40 m? = 26.7 m-1 Focal length is calculated as follows, 1x100 cm 26.7 = 3.74 cm

E) 1 - 1 - 1 = +- v=fu f +u (3.74 cm)(4.1 cm) 4.1 cm+(3.74) =1.96 cm Image formed is real. Magnification is calculated as follows, m=-V -1.96 4.1 = -0.48 Image is inverted and smaller in size, which is formed on the other side of the lens.