Tricky Path Difference Problem in waves, help (attachment)

how should i do this problem?

  • using visual interpretations and say that it follows logically

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  • using equations to analyse it

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ihatephysics0
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Tricky Path Difference Problem in waves, help! (attachment)

Homework Statement



check the attachment

Homework Equations



none needed for part (d)

The Attempt at a Solution


the first three questions of part (c) i got them out. answers were

(c) (i) 2x10 ^8 ms–1
(ii) 3.33333x10^–7 m
(iii) 30

BUTTTT i can't seem to understand how to do both questions in part (d). Any help?? please!
Any thorough and logical reasoning with answer would be best as i need to understand how to do these questions (before exams come!). thanks :)
 

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  • Physics exam - 2003_pagenumber.001.png
    Physics exam - 2003_pagenumber.001.png
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haruspex said:
Isn't c iii also the answer to d i?
For d ii, what affect has the plastic had on the path length through that slit to any given point on the screen?
er... to be honest I'm really unsure about this question, it's still vague. Can you please provide some reasoning to why it should be the same for d (i) ? and yes i know the effect but i cannot arrive at a definite and certain conclusion for d (ii)?

the answer for d (i) says
"The plastic region would normally have 20 wavelengths across it, but when present 30
wavelengths would exist. Hence 10 extra wavelengths have been introduced by using the
plastic, so p.d. = 10" i still need clarification on this...
 


ihatephysics0 said:
the answer for d (i) says
"The plastic region would normally have 20 wavelengths across it, but when present 30
wavelengths would exist. Hence 10 extra wavelengths have been introduced by using the
plastic, so p.d. = 10" i still need clarification on this...
Ah yes, I forgot to subtract the number of wavelengths for the airspace replaced by the plastic. So indeed: 30 through the plastic, 20 through the corresponding section of path through the other slit: resulting difference for a point equidistant (as measured by a ruler) from both slots = 10 wavelengths.
That said, there is a possible ambiguity in the question. It refers to "the centre of the fringe pattern". Now, is that the centre of the pattern as it was without the plastic (or equivalently, the centre of the screen as defined by the positions of the slits), or is it the new centre of pattern based on how the fringes now appear? I think they mean the old centre, but it should be clearer.
OK, so it's added 10 for the path to the centre, so what has it done to paths to other points on the screen?
 


haruspex said:
Ah yes, I forgot to subtract the number of wavelengths for the airspace replaced by the plastic. So indeed: 30 through the plastic, 20 through the corresponding section of path through the other slit: resulting difference for a point equidistant (as measured by a ruler) from both slots = 10 wavelengths.
That said, there is a possible ambiguity in the question. It refers to "the centre of the fringe pattern". Now, is that the centre of the pattern as it was without the plastic (or equivalently, the centre of the screen as defined by the positions of the slits), or is it the new centre of pattern based on how the fringes now appear? I think they mean the old centre, but it should be clearer.
OK, so it's added 10 for the path to the centre, so what has it done to paths to other points on the screen?

ah i think we are finally on the same wavelength (giggle), it makes sense now. thank you so much! if i could hug you i would! and the last question is pretty easy to answer, thanks again!