QuestionDetails: Tags: Physics , College Physics(8th) , Algebra BasedPhysics Tag

Question

QuestionDetails: Tags: Physics, College Physics(8th), Algebra BasedPhysics Tags: Physics, College Physics(8th), Algebra BasedPhysics Tags: Physics, College Physics(8th), Algebra BasedPhysics
I know there was an earlier post on this but I still didn'tget it right. Can anyone help? Thanks! QuestionDetails: An observer to the right of the mirror-lens combination shown inthe figure below sees two real images that are the same size and inthe same location. One image is upright and the other is inverted.Both images are 1.50 times larger thanthe object. The lens has a focal length of 10.0 cm. The lens and mirror are separated by 40.0cm. Determine the focal length of the mirror. (Don't assume thefigure is drawn to scale.)
fmirror = 1
Your answer differs from the correct answerby 10% to 100%. cm

Explanation / Answer

   focal length of the lens f1 = 10.0 cm    for the given probloem let p1 bethe object distance and q1 is the image distance    the magnification is 1.50    the magnification M is given by    M = - (q1 /p1)        = - 1.50    q1 = 1.50 p1    the thin lens equation is given by    (1 / f) = (1 / p1) + (1 /q1)    (1 / 10.0 cm) = (1 / p1) + (1 / 1.50p1)    p1 = ......... cm    in the figure the upright image is formed bythe light that passes through the lens after reflecting    from the mirror    now the object for the lens in this uprightimage formation is the image formed by the mirror    so to get an upright image at the samelocation as the inverted image the image formed by the    mirror must be located at the position ofthe original object so the object and image distance for    the mirror are equal and their common valueis    qmirror - pmirror =40.0 cm - p1                           = ....... cm    using the mirror equation we get    (1 / pmirror) + (1 /qmirror) = 2 / R                                           = (1 / fmirror)    solve for the focal length of the mirror    fmirror = ......... cm                           = ....... cm    using the mirror equation we get    (1 / pmirror) + (1 /qmirror) = 2 / R                                           = (1 / fmirror)    solve for the focal length of the mirror    fmirror = ......... cm

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.