I\'m having trouble just drawing it out. I would appreciate thesteps as explicit

Question

I'm having trouble just drawing it out. I would appreciate thesteps as explicitly as you can. Thank you!
An object, 0.75 cm tall, is placed 12.0 cm to the left of adiverging lens (f = -.8.00cm). A converging lens is placed 8.00 cmto the right of the diverging lens. The final image is virtual andis 29.0 cm to the left of the fiverging lens. Determine (a) thefocal length of the converging lens and (b) the height of the finalimage. I'm having trouble just drawing it out. I would appreciate thesteps as explicitly as you can. Thank you!

Explanation / Answer

An object, 0.75 cm tall, is placed 12.0 cm to the left of adiverging lens (f = -.8.00cm). A converging lens is placed 8.00 cmto the right of the diverging lens. The final image is virtual andis 29.0 cm to the left of the diverging lens. Determine (a) thefocal length of the converging lens and (b) the height of the finalimage. initial object distance=u=12cm f= -6cm so 1/v=-1/8-1/12=>v=-4.8cm now the object distance with respect to the converginglens=12.8cm image distance=29+8=37cm so f=37*12.8/49.8=9.51cm hence the focal length of the converging lens = 9.51cm. (b) the height of the final image. magnification due to diverging lens and converging lens           =[4.8/12]* [37/12.8]=1.156 height of the final image = 1.156*0.75=0.8675cm

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