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AS Physics P1 MCQs Preparation Thread.

Obsidian Fl1ght

Obsidian Fl1ght

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Sajeeb Chowdhury said:
could ny 1 please solve this.... :)

View attachment 28839
Click to expand...
d sin thetha = n lamida
d = 1/n, n = lines/m
So n = 500 * 1000 lines/m
d = 1/n = 1/500,000
thetha take 90
lamida is 600 * 10 ^-9 m

Find n.
n is 3.33 ie 3 complete images on one half of screen.
On both halves = 3 * 2 = 6
But u also have to consider central maxima.
So total maximas (images) is 6 + 1 = 7
 
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6Astarstudent

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Sajeeb Chowdhury said:
could ny 1 please solve this.... :)

View attachment 28839
Click to expand...
use the equation nλ=dSinθ where n(number of order), λ(wave lengeth), d(distance between each grating) and θ being the angle.
since he's looking at the maximum number of images so Sinθ= 1. and nλ=d.
to find d, you divide 1mm/500 so you get 2x10^-6/
n=d/λ=2x10^-6/(600x10^-9)= 3.33 on each side of the zero order
so you add 3+3 to find images on both sides of Y and add the zero order line which also forms an image. so 3+3+1=7

oops somebody already answered
 
Obsidian Fl1ght

Obsidian Fl1ght

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Champ101 said:
lol sorry for that. Good luck for the exam today hope you do very well :D
Click to expand...
Haha no harm done.
U admitted ur mistaken-ness and that, my friend, means u've got guts.
:D
Technnically speaking, everyone here HAS guts... but yeah u get the point.
U too!!!
 
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Abu mota said:
answer is D, because only points within the overlap are in phase and so A and C are wrong. also B is wrong because stationary waves have varying amplitudes.
Click to expand...

thanks
so u mean to say from N1 TO N2 point ass in phase then from n2 to n3 is it?
 
Obsidian Fl1ght

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Abu mota said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf Q.27 q.37,q.40
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w07_qp_1.pdf Q.29,q.35,q.40
Click to expand...
27 shd b A.
No. 1. Nodes are always formed at closed ends so discard D.
No.2 In one loop, ie between two nodes, particles are in phase. WITHIN one segment. (Or loop)
So particles r in phase, one direction, in bottom segment.
Particles are in phase in second (top) segment. BUT particles in bottom and top segments ARE NOT in phase with EACH other.
I hope u get what I mean.
Amplitude of particles equidistant from a node is same.
So A.
 
Abu mota

Abu mota

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Obsidian Fl1ght said:
27 shd b A.
No. 1. Nodes are always formed at closed ends so discard D.
No.2 In one loop, ie between two nodes, particles are in phase. WITHIN one segment. (Or loop)
So particles r in phase, one direction, in bottom segment.
Particles are in phase in second (top) segment. BUT particles in bottom and top segments ARE NOT in phase with EACH other.
I hope u get what I mean.
Amplitude of particles equidistant from a node is same.
So A.
Click to expand...

yes, i understood. thank you.
 
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