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    https://5505b400f275fd9f29b9b4679ca5...%20Physics.pdf

    Please help me with Question 7 part C... I cannot understand how to calculate that without knowing the refractive index of the liquid. :/ Is there something I'm missing?
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    (Original post by Someboady)
    https://5505b400f275fd9f29b9b4679ca5...%20Physics.pdf

    Please help me with Question 7 part C... I cannot understand how to calculate that without knowing the refractive index of the liquid. :/ Is there something I'm missing?
    Read the last line of the question
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    At this point the angle of refraction is 27º and the ray is totallyinternally reflected at P for the first time.

    "for the first time".... does that mean that the critical occurs when the angle of refraction is 27 degrees :/ ?

    (Original post by Spelly456)
    Read the last line of the question
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    (Original post by Someboady)
    At this point the angle of refraction is 27º and the ray is totallyinternally reflected at P for the first time.

    "for the first time".... does that mean that the critical occurs when the angle of refraction is 27 degrees :/ ?
    No, you need to use a triangle to find the critical angle
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    (Original post by Someboady)
    At this point the angle of refraction is 27º and the ray is totallyinternally reflected at P for the first time.

    "for the first time".... does that mean that the critical occurs when the angle of refraction is 27 degrees :/ ?
    Think about which angle you will be measuring as the critical angle when you consider the water-glass boundary. You're almost right!
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    So what is the calculation required for the answer? its only a 1 marker? lol
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    (Original post by hellomynameisr)
    So what is the calculation required for the answer? its only a 1 marker? lol
    90-27 = 63 degrees
    That is the answer
    Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle
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    (Original post by Spelly456)
    90-27 = 63 degrees
    That is the answer
    Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle
    Oh right, I see. Thanks
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    (Original post by Spelly456)
    90-27 = 63 degrees
    That is the answer
    Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle

    still slightly confuzzled sorry.... So I've drawn a new diagram... part a says to continue the ray that's been totally internally reflected..So I drew the ray reflected off the bounday and coming out of the glass. I didn't do any calculations.

    Am I correct in saying the ray of light, when striking the boundary between liquid and glass, strikes at the angle of 63 degrees... this can be figured out using 90-27 = 63.... But I don't understand how that is the critical angle. My understanding of the critical angle is that the angle of refraction must be 90 degrees.. Does this mean my diagram in part A is wrong and that the ray should continue at 90 degrees from the normal and continue travelling along the boundary :/
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    (Original post by Spelly456)
    90-27 = 63 degrees
    That is the answer
    Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle
    I just read the examiners report :
    For part (c), candidates needed to realise the incident angle had just passed the critical angleand therefore the critical angle would be 63° to two significant figures.

    Why has it "just" passed the critical angle... why can't it have passed the critical angle by a large amount? what tells us that its only passed the critical angle "slightly"?
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    (Original post by Someboady)
    I just read the examiners report :
    For part (c), candidates needed to realise the incident angle had just passed the critical angleand therefore the critical angle would be 63° to two significant figures.

    Why has it "just" passed the critical angle... why can't it have passed the critical angle by a large amount? what tells us that its only passed the critical angle "slightly"?
    In the question it says the ray is totally internally reflected for the first time so 63 degrees is just after the critical angle. You are right that at the critical angle the angle of refraction is 90 degrees so the critical angle would be slightly below 63 degrees.
    Hope this makes sense.
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    (Original post by Someboady)
    I just read the examiners report :
    For part (c), candidates needed to realise the incident angle had just passed the critical angleand therefore the critical angle would be 63° to two significant figures.

    Why has it "just" passed the critical angle... why can't it have passed the critical angle by a large amount? what tells us that its only passed the critical angle "slightly"?
    The reason you quote the answer as 63 degrees is due to rounding of 62.99999
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    (Original post by Spelly456)
    In the question it says the ray is totally internally reflected for the first time so 63 degrees is just after the critical angle. You are right that at the critical angle the angle of refraction is 90 degrees so the critical angle would be slightly below 63 degrees.
    Hope this makes sense.
    Yep it makes sense thanks! However when it comes to generalising... If a similar question was to come up and it stated that it was "totally internally reflected for the first time"... would this again mean that the critical angle is slightly below the angle of incidence (in ANY situation)?
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    (Original post by Someboady)
    Yep it makes sense thanks! However when it comes to generalising... If a similar question was to come up and it stated that it was "totally internally reflected for the first time"... would this again mean that the critical angle is slightly below the angle of incidence (in ANY situation)?
    Yes it would do
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    (Original post by Spelly456)
    Yes it would do
    That's great Thanks a bunch dude!
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    (Original post by Someboady)
    That's great Thanks a bunch dude!
    No problem, happy to help!
 
 
 

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