Question

Select Diverging lens Knowing that the scale of the ruler in the simulation is in centimeters, set the focus length of the diverging. lens to 2.5 cm and the high of the object to 2 cm approximately. 1. Locate the object at the left of diverging lens. a. Print the simulation and draw the third principal ray to form accurately the image of the object. b. Where exactly the image of the object will be formed? hint - watch your units 1 1 1 - + - paf C. Is the image real/virtual? d. Is the image upright/inverted? e. What is the magnification of the lens? m2 = - ho P f. Calculate the high of the image.

Answer 1

focal length of the divenging lens, f= -2.5cm Height of the object, ho=2cm Let the object be placed at som infornt of the object distance, p=-5.0cm lens, 1 F. I q f = 21+ f -- => 9= - = -1.670 2 p -2.5 -5 (bl 11-1 f 9 $> Р distance of the image, q= -1-67 cm (The image is virtual (d) The image is upright q (e) magnification of the lens, m= (1) magnitication, ma Height of the image, hi = 0.66cm - 167 -5 => m=0.33 P hi q =) hi hox q 2x0.33=0.66cm こ一 ho Р Р