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Simple optics

  1. Dec 6, 2007 #1
    Question:
    At the bottom of a glas bath lies a mirror covered by 20cm of water.In the air,30 cm above the water there hangs a lamp.Find the distance from the mirror at which an observer looking into the water will see the image of the lamp in the mirror

    Relevant Formula:
    apparent shift [tex]\Delta t=t(\frac{\mu}{1}-1)[/tex]

    Attempt:
    the apparent shift of the lamp is 10 cm so the situation is same as a lamp placed in the system without water at a distance of 60cm from the mirror.
    But that's all i could get i dont understand the question well .
     
  2. jcsd
  3. Dec 7, 2007 #2

    rl.bhat

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    Refractive index = Real depth / apparent depth.
    Find the apparent depth of the lapm in the mirror. From this find the apparent distance of the lamp from the mirror. Same will be the distance of the image in the mirror.
     
  4. Dec 7, 2007 #3
    but that doesnt give the answer answer is 45cm
     
  5. Dec 7, 2007 #4

    rl.bhat

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    4/3 = 20/apparent depth
    Apparent depth = 15 cm.
    Now apparent object distance = 30 + 15= 45cm = Image distance
     
  6. Dec 7, 2007 #5

    Shooting Star

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    Hi Bhat, I've got some questions. I arrived at the answer through more complicated reasoning. How come your simple formula gives the same answer?

    If you look at the lamp from inside the water, the apparent distance from the interface should be (4/3)30 cm = 40 cm. (More than actual.) From the mirror surface, it should look as if the lamp is at a dist of 20+40=60 cm. So, the image of this should form at a dist of 60 cm behind the mirror.

    Now if you look from the air, the actual depth of image is 20+60=80cm. So, apparent depth should be (3/4)80=60 cm from the air-water interface. But the mirror is at an apparent depth of (3/4)20=15 cm. So, the distance of the image from the mirror according to an observer looking into the water should be 60-15=45 cm.

    What I’m asking is what exactly is the correlation between your and my methods?
     
  7. Dec 7, 2007 #6
    thanks shooting star and rl.Bhat.
    i got shooting star's method
     
  8. Dec 7, 2007 #7

    rl.bhat

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    If you look at the lamp from inside the water, the apparent distance from the interface should be (4/3)30 cm = 40 cm. (More than actual.) From the mirror surface, it should look as if the lamp is at a dist of 20+40=60 cm. So, the image of this should form at a dist of 60 cm behind the mirror. For a fish it is true.
    Now if you look from the air, the actual depth of image is 20+60=80cmThis in not correct. Because you are in air and object is in air. Between mirror and you water is there. Therefore the actual distance of the object from the mirror cannot be more than 50cm.
     
  9. Dec 7, 2007 #8

    Shooting Star

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    The mirror surface is a fish, as far as this problem is concerned. If we draw a diagram, we'll see that the light rays appear to diverge at a dist of 60cm from the mirror surface.

    The actual dist doesn't count, it's the dist which the mirror surface perceives the light to be coming from, viz., from a dist of 60 cm.

    Tell you what, I'll formulate the problem in terms of symbols. The water thickness is w, the lamp is d dist above it, the ref index of water is n. Now you tell the ans according to your method, and I'll do it by my method.

    I just won't believe that the two similar answers we got are co-incidences, especially when I think my logic is fine. Maybe a good diagram will resolve it immediately. I'll do it when I'm free, maybe after 8-10 hrs. Best wishes.
     
  10. Dec 7, 2007 #9

    Shooting Star

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    All right. Listen, the two methods are absolutely identical, and in this problem gives the result d + w/n.

    In your scenario, the water is replaced with air, thus bringing the mirror closer to the lamp by w/n, making the effective distance between the lamp and mirror surface d + w/n. Naturally, the image is at the same dist behind the mirror.

    In my method, I was replacing the air by water, thus making the distance of the lamp further from the mirror. But later I had to revert back to air when observing the dist of the image from the mirror with the observer in the air. Ultimately both gives the same result. Skip the derivation. It's elementary and I'm too tired to write it now.

    Your method is much simpler for this kind of problem, as is evident, provided you explain it to the newbies.

    Cheers.
     
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