Question

2) A diverging lens is placed 20. cm away from an object, 80. cm from the diverging lens is a converging lens whose focal length is 25 cm. 35. cm from the converging lens is a screen showing an image of the object. a) What is the focal length of the diverging lens? f = 25. cm o *20. cm → 80. cm + 35 cm → b) What type of image is it? Circle one Real or Virtual c) What type of image is it? Circle one Upright or Inverted d) If the original object is 4.1 cm high how tall is the image?

Answer 1

for converging lens: f=25cm V=Brom object location =-U2 I f ottby et V - U2 t bo to 35-25 Jo 25 35 25x35 25x35 You go 25X35 87.5 cm location of object for ati converging leus os actually location of image formed diverging uns: by Converging en 7.5cm دے دیا 87.5cm image location of diverging lens is Van Via-Tosem

U = -20cm fa? ㅗ Vi L U ㅗ f ㅗ -7.5 - 7.5-20 a -20 f 20 I 7.5 20x7.5 f - 12.5 2047.5 f. 20x7.5 12.5 = -12 cm (b) on screen so (a) focal length image es oby obtained èt ies real are always inverted. ho magnification from diverging lens: Real images 4.1cm mo Vi VI -P.S - 20 7.5 20 image height from diverging lensa ho ho 7: - hi 20 9.5 X 4.1 1.5375 cm magnification from converging deus: 105. m₂ = V2 Us - 35 ha he -87.5 1.5375 = 0.615cm. 1.5375835 87.5