3. Explain using mathematics and a geometric diagram the difference between converging and diverging lenses. [5 points} 2. Choose three of the above lens problems and draw a scale diagram for each one that shows geometrically how the distance to the image is found. (5 points each} Be sure to: 0 Label 50, s.', and f clearly. 0 Use a ruler or straight edge for neatness 0 Draw to scale. 1. Find the distance to the image in the following lens problems and state whether the image is virtual or real AND whether it is upright or inverted. (4 points each) a. An object is placed 40. mm away from a diverging lens with a focal length of 10. cm. do = 40 mm /- 10cm = 1/40mum + 1/di distance = - 20 cm P = - 10 cu 1/di = 1/- 10 cm - 1/40 mm virtual 1/ P = 1/ do + 1/ di di= - 2.0om upright b. An object is placed 2.5 cm away from a converging lens with a focal length of 1.5 cm. do = 2.50m 1/1. sem = 1 /2.sam + 1 / di distance = 6.0cu f = 1.5 am virtual YP - Ydo + 1 / di 1 / di = 1/1.50m - 1/ 2. 50 M upright di = 6.Damn c. An object is placed 7.0 cm away from a converging lens with a focal length of 7.0 cm. do = 7 0cen f = 7 . 0cen ' / 70cm = 1 17.0cm + " / di distance = 14.0 cum real 1/ p = 1/d + 1/di 1/ di = 1/7.0 cm - 1/2.Ocm inverted di = 14.ocm d. An object is placed 65 mm away from a diverging lens with a focal length of 30. mm. do = 65 mm 1/- 30 mm = 165mm + 1/di distance = - 4.5 cm f = - 30 mm 1/ di = 4- 30 mm - 1/65 mm virtual upright 1/p = Ydo+ 1/di di = - 4.5 cm e. An object is placed 12 cm away from a diverging lens with a focal length of 6.0 cm. do = 12 cm 1-6.0cen = 1/ 12cut 1/ di f = - 6.0 cuu distance = - 4.0cm virtual / f = 1/do +"/di 1 /di = 146.0cm - 1/12cm upright di = - 4.Ocm