Physics P2 Help Required

[MAVtKnmJ]

I have a question.

W07 P2 4 (c) - I've got the area to be 2.0 x 10^-6 but I don't understand why we're subtracting this from 3.2 x 10^-6.
4 (d) - I don't exactly get the logic behind the mark scheme's answer!

Yeah, I didn't get that one either. But what I have derived from it is that, since the question asks for the "Maximum" area, then it means the cross-section must be divided into two circular portions. One "Minimum" area which we calculated and one "Maximum" area which we get by subtracting the minimum area we found after stretching from the original area.

 

[mz23]

june 09, paper 21, question 5,b...please help asap!

 

[Azazel]

june 09, paper 21, question 5,b...please help asap!

It's been answered already. Turn a page back.

 

[MAVtKnmJ]

It's the same thing: try understanding what the equation means. n is the point where displacement is max, if you add half a wavelength to it, you'll be where displacement is zero. If you remove half a wavelength, you'll still be where the displacement is zero.

I still don't understand what you mean.
I've been taught that the formula is (n - 1/2) (lambda) but if I use that in this question my answers don't match up.
Now when I use (n + 1/2) lambda then answers match up.

So why is that? I don't think both are the same things!

 

[mz23]

ok thanx but what if we use (2n+1)(lambda)/2? and i've forgotten what the n stands for?

 

[Azazel]

(2n+1)(lambda/2) = [(2n+1)/2] * lambda = (n + 1/2) * lambda

Same thing. As for n, I believe it's the number of anti-nodes (i.e max displacement). I'm not entirely sure though, someone else care to reply?

 

[MAVtKnmJ]

n = the number of maximas or minimas produced or the Order Number in diffraction grating formula!

Can you reply to my last post?

 

[Azazel]

Hm, you're right!

I just re-checked beacon_of_light's solution again and his/her answer seems to be wrong.

I quote:
------------
n+(0.5)(0.33) = 0.28 ....and n = 0.35..

n+(o.5)(0.0825)= 0.28...and n= 3.39...
------------

The first part is correct, (n+0.5)(0.33) = 0.28, n + 0.5 = 0.28/0.33, n = 0.35

As for the second part, (n+0.5)(0.0825) = 0.28, n + 0.5 = 0.28/0.0825, n = 2.89, NOT 3.39.

If we follow on, he/she states that odd numbers i.e n=1 and n=3 are to be counted. This is, clearly, no longer possible.

I suggest you follow sizbeauty's solution. It seems to be clear.

In any case, I'm just throwing a thought out on this one. To me, logically, n + 0.5 and n - 0.5 should give the same answer. If we had used n - 0.5, we'd get 1.35 and 3.89. The whole numbers between are 2 and 3 (i.e TWO minima). If we used n + 0.5, we'd get 0.35 and 2.89. The whole numbers between are 1 and 2 (i.e TWO minima).

 

[intel1993]

yes tthe solutions seems to be wrong as i have aslo taught that for destructive interference ptah difference is equal to ( n- 1/2 lamda)

 

[beacon_of_light]

guys i am sorry ive left sum rome of confusion........use (n+1/2) x lambda...keep in notice the brackets!!!u'll surely get the right ans

 

[Azazel]

I'm sorry, beacon_of_light, but I already assumed that you wanted to say (n + 0.5)*lambda.

You still get the incorrect answer.

 

[intel1993]

yes because......................path difference is (n-1/2)*lambda for destructive interference........................

 

[beacon_of_light]

I'm sorry, beacon_of_light, but I already assumed that you wanted to say (n + 0.5)*lambda.

You still get the incorrect answer.

No the path differnce for destructive interference is (n+1/2)x lambda...check out any book if u want...longman physics

 

[Azazel]

My point was that the answer is incorrect. As I said earlier, perhaps you don't need to take odd numbers i.e n = 1, n = 3, but you need to take /any/ whole number e.g in this case, n=1 and n=2.

 

[beacon_of_light]

no when we are considering minimas....odd intergral multiples of wavelengths are considered....

 

[Azazel]

but you can't get the correct answer if you consider the odd integrals.

I guess this is the only CORRECT way to solve it for now:

viewtopic.php?f=26&t=1332&start=50#p20445

 

[ashhadazam]

Which question u talking about ??? it doesnt matter if u take n-1 or n+1.. U should get your path difference as 0.5, 1.5, 2.5..i.e, middle values.. Where N is the first antinoadal point i think :O

 

[Amanina]

hey guys. can anyone help me with this question.

MJ06 Q7 b.

why is the answer state that 'shorted lamp A'. i don't understand.

 

[musa]

Bhai will anyone explain me jun 09 ques 5 Variant 1. Its really tough yar....

 

[beacon_of_light]

No what is the problem actually, u can't get to the right values of 'n' or what?