Correct, computer gets: 37.3cm 2). If you place an object very far from the mirr

Question

Correct, computer gets: 37.3cm

2). If you place an object very far from the mirror, what is thelocation of the image?

Correct, computer gets: 37.3cm

3. What is the magnification of the image?

Correct, computer gets: 0

Answer:

5. What is the magnification of theimage?

Answer:

6. If a diverging lens of focal length -67.0 cm is then placed27.0 cm from the first lens on the opposite side of the lens fromthe object, what is the location of the new image?

Answer:

7. What is the magnification due to thetwo-lens system?

Answer:

Explanation / Answer

(4)              Given :                     Focal length (f) = 30.0   cm                     object distance ( u) = 47.65 cm We know that :                        1/ f = 1/ u   + 1/ v             or        1/ v = 1/f - 1/ u                                =   1/ 30 - 1/ 47.56         Imagedistance (v) = 81.25 cm (5)                     Magnification (M) = - v / u                                                  = - 81.25 / 47.65                                                  = 1.705 (6)                 Here Object distance is :   81.25 -27.0   cm                                                                u`   = 54.25   cm                           Focal length (f ` ) = - 67.0 cm                We have :                              1 / f ` = 1/ u` + 1/ v`                   or        1/ v`  =   1/ f ` - 1 / u`                                         = - 1 / 67.0 - 1 / 54.25               New Image location is :                                  v` =   - 29.977   cm                                        ˜  - 30 cm (7)               Magnification (M `) = - v` / u`                                              = 30.0 / 54.25                                               = 0.553 Hope this helps u!

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