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Physics: Post your doubts here!

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Hi everyone, AsSalamoAlaikum Wr Wb...

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your PHYSICS doubts in this thread. InshaAllah other people here will help me and you all. :D ;)

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites! :)
Thanks!
Jazak Allah Khair!


Physics Notes:

Some links & Notes - by destined007

As physics p1 MCQS YEARLY Solved [explaination]
Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

Physics Summarised Notes (Click to download)

AS and A-Level Physics Definitions

A2: Physics Revision notes - by smzimran

Paper:5 Finding uncerainty in log - by XPFMember

Physics Paper 5 tips - by arlery

Physics Compiled Pastpapers: <Credits to CaptainDanger for sharing this..>

Here are the compiled A level topical Physics questions in PDF form...

Paper 1 : http://www.mediafire.com/?tocg6ha6ihkwd

Paper 2 & Paper 4 : http://www.mediafire.com/?g65j51stacmy33c

(Source : http://www.alevelforum.com/viewtopic.php?f=15&t=14)
 
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I'm assuming it's a diffraction grating question as you have cut off the first part of the question AND the paper number -.-

Second order maxima, thus n = 2
substituting in dsin x = n times wavelength,
you get d.
now d is the fringe spacing. The number of fringes in 1m is 1/fringe spacing.

I'm sorry I can't exactly explain why but remembered something I learnt in primary school.
Like if 1 box is 5 cm, and a large box is 100 cm, how many boxes can fit in a large box or something like that...
 
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I'm assuming it's a diffraction grating question as you have cut off the first part of the question AND the paper number -.-

Second order maxima, thus n = 2
substituting in dsin x = n times wavelength,
you get d.
now d is the fringe spacing. The number of fringes in 1m is 1/fringe spacing.

I'm sorry I can't exactly explain why but remembered something I learnt in primary school.
Like if 1 box is 5 cm, and a large box is 100 cm, how many boxes can fit in a large box or something like that...
Yeah if one fringe spacing occupies x mm(also the fringe spacing cause fringe spacing is from the center of one fringe to the center ifvanothervif I'm not wrong)) then how many fringes will occupy N mm
 

zem

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I'm assuming it's a diffraction grating question as you have cut off the first part of the question AND the paper number -.-

Second order maxima, thus n = 2
substituting in dsin x = n times wavelength,
you get d.
now d is the fringe spacing. The number of fringes in 1m is 1/fringe spacing.

I'm sorry I can't exactly explain why but remembered something I learnt in primary school.
Like if 1 box is 5 cm, and a large box is 100 cm, how many boxes can fit in a large box or something like that...
Oh i see thankyou!
 
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Guys..
The conditions for the *young double slit experiment * and for the *interference pattern * are the same right.. Confused :$

And these are the conditions ..
-> monochromatic light
-> coherent sources
-> same frequency and amplitude
-> plane of polarization should be the same


Please add more .. If required and correct if any wrong :)
Thank u :D
 

zem

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Yeah if one fringe spacing occupies x mm(also the fringe spacing cause fringe spacing is from the center of one fringe to the center ifvanothervif I'm not wrong)) then how many fringes will occupy N mm
And could you help me with that second questions b part?
 
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How do we calculate resultant intensity?
Here is the question
The intensity of wave A at point P is I. Intensity of wave B at point P is (4/9)I. The phase difference between these two waves 180 degrees. Wavelength of A and B is 3x10^-4 and 2x10^-4 cm respectively. Frequency of both waves is "f". What is the resultant intensity at P in terms of I?
 
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Well that's the strength part.. The closer the field lines the stronger the strength..
So that gotta be on the top.. Where they r close..
And constant is when the have equal distance btw them as in equal.. Strength :)

thanks alot for an answer :)

But can you please elaborate on how to find the decreasing field strength part?

Thanks again :)
 
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