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Refraction through a glass slab + reflection off a concave mirror

  1. Apr 8, 2015 #1

    Suraj M

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    1. The problem statement, all variables and given/known data
    An object is placed 21 cm in front of a concave mirror of radius of curvature of 10 cm. A glass slab of thickness 3 cm and RI =1.5; is then placed close to the mirror. Find the position of the final image.
    Take distance of closer side of block as 1 cm from the mirror.

    2. Relevant equations
    ##\frac{1}{f}= \frac{1}{u} + \frac{1}{v}##
    u is the object dist. and v is the image dist.
    ##lateral~ shift~=~t\frac{\sin(i-r)}{\cos(r)}##

    3. The attempt at a solution
    Ignoring the block i get ##v = -6.56 cm##
    I don't know how to consider the block... i tried this..
    WIN_20150408_190509.JPG
    i need the length of B'B'₂
    I have the measure of a perpendicular drawn from A' to the line through A'₂ which was suppose to go through the focus..
     
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2
    First consider a point object on the principal axis for sake of simplicity . Second ,forget about the mirror for a while . The ray from object undergoes twice the refraction from the slab .

    Can you calculate the normal shift produced by the slab ?
     
  4. Apr 8, 2015 #3

    Suraj M

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    But how, doesn't that depend on the angle of incidence.
    Which ray?
    passing through the centre of curvature →normal shift = 0 as it is refracted twice ..
     
  5. Apr 8, 2015 #4
    Forget the mirror.

    I suggested you to consider a point object placed at __ cm from the slab . Draw any ray from the object at any angle from the priniple axis . Angle is not important .Center of curvature is not involved here .
     
    Last edited: Apr 8, 2015
  6. Apr 8, 2015 #5

    Suraj M

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    ##lateral ~shift = t \frac{\sin(i-r)}{\cos(r)}##
    so the lateral shift does depend on the incident angle, doesn't it?
    but as you said if i take the ray at an angle(##=i##) to the principal axis then
    lateral shift = ##3\sin(i) - 4.5 \cos(i) ##
     
  7. Apr 8, 2015 #6
    Normal shift and lateral shift are two different things . We are interested in calculating normal shift not lateral . I have no idea about the formula you have mentioned . Please check your notes on how to calculate the normal shift produced by the glass slab .Again , normal shift doesn't depend on the angle .

    Have a look at the picture attached .

    You need to calculate distance OA .
     

    Attached Files:

    Last edited: Apr 8, 2015
  8. Apr 8, 2015 #7

    Suraj M

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    I thought we needed lateral shift.. ok normal shift ..then its just 5.67 cm (##\frac{d}{n}##)
    but i still think its lateral shift and not normal shift, normal shift would be this.. Snapshot.jpg
    I think the diagram you have drawn is actually for lateral shift as the emergent ray is parallel to the incident ray...
    So i cannot apply the formula ##d' = \frac{d}{n}##or normal shift = ##d(1-\frac{1}{n})## where n is RI of glass wrt air
     
  9. Apr 8, 2015 #8
    Don't you think the emergent ray would be parallel to the incident ray after two refractions due to the glass slab ?

    This is the right formula . So , now what is the normal shift ?

    However what is the answer given to the problem?
     
  10. Apr 9, 2015 #9

    Suraj M

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    The answer given is 7.67 cm.
    Yes it will be, so if the emergent ray is parallel to the incident ray... there is a lateral shift, not normal shift, right?? A component of the lateral shift is the distance between the position and apparent position.
     
  11. Apr 9, 2015 #10
    Ok...I am getting this value .

    Calculate the value of normal shift . Then , find the apparent distance of the object from the mirror . Now , use the mirror formula .
     
  12. Apr 9, 2015 #11

    Suraj M

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    wait should i do this??
    ##d' = \frac{17}{1.5}## then ##d' + 3 + 1 = u##??
    then i get 7.42 cm
    is that right?
     
  13. Apr 9, 2015 #12
    Sorry , I don't understand your working .

    1) What value do you get for normal shift ?
    2) What is the apparent distance of the object from the mirror ?
     
  14. Apr 9, 2015 #13

    Suraj M

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    for (1) i wanted to know if i should take the distance between the glass block(17 cm ) or the distance from the mirror..
    anyway i get ##d' = 11.33 cm## and apparent object distance as 15.33 cm
     
  15. Apr 9, 2015 #14
    This is incorrect .

    Normal shift = ##(1-\frac{1}{\mu})t## , where ##\mu## is the refractive index and ##t## is the thickness of the glass slab .

    Now please answer the two questions I asked in post#12 .
     
  16. Apr 9, 2015 #15

    Suraj M

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    ok is the normal shift 5.66 cm ??
    if so then apparent distance is 17 - 5.66 = 11.33 cm
    which is the same as ##\frac{17}{n}##
     
  17. Apr 9, 2015 #16
    No

    No

    I have given you the formula in post #14 . Please use it .
     
  18. Apr 11, 2015 #17

    Suraj M

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    I wasn't sure what you were trying to do here, but i think i get the idea..
    you're trying to calculate the effective object distance which would become ##(u=21 -1 = 20 cm )##
    so then applying mirror formula i get v = 6.67 cm ... but due to the slab there is a second refraction causing the image to shift by 1 cm ... = 6.67+1= 7.67 cm
    I'm sorry it took me so much time.. :) and for irritating you so much...
    Thanks a ton!
     
  19. Jun 12, 2016 #18
    What if we place the glass slab at 5 cm from the mirror? Will the image be still formed at 7.67 cm? Or will there be some change. In my opinion there shouldn't be any but please can anyone confirm?
     
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