(a) How far from a 52.2 mm focal-length lens must an object be placed if its image is to be magnified 2.40 x and be real?
(b) What if the image is to be virtual and magnified 2.40 x ?
2007-02-02 13:23:35 · 3 answers · asked by Bob B 2 in Science & Mathematics ➔ Physics
The thin lens formula is 1/f = 1/Do + 1/Di
where f = focal length, Do = object distance, and Di is image distance.
Magnification is defined to be M = Hi/Ho
where Hi = image height and Ho = object height
But magnification is ALSO = - Di/Do
Real images formed by converging lenses (with + focal lengths) are inverted, or have negative magnifications. (If you wonder how I know this, google "optics principal ray diagrams" and you will see how to use pictures to analyze problems like these.)
So,
(1) -2.4 = -Di/Do
(2) 1/52.2 = 1/Do + 1/Di
And now you've got two equations and two unknowns! (Solve Eqn 1 for Di and then substitute that result into equation 2, and you'll get the answer to question a.
(b) If the image is Virtual, the magnification is positive, so the magnification equation for part b becomes
(1) 2.4 = - Di/Do
(2) Equation 2 is the same as eqn 2 above.
(For the virtual image, you should find that the image distance is negative if you solve the system of equations for Di.)
2007-02-02 14:40:15 · answer #1 · answered by Dennis H 4 · 0⤊ 0⤋
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2016-09-28 08:34:39 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Sorry after reading a few of your answers, I realized you are quite a dikhead. So there: no answer for you silly kid.
2007-02-02 13:33:32 · answer #3 · answered by catarthur 6 · 0⤊ 0⤋