ALL CIE PHYSICS DOUBTS HERE !!!

[Deadly_king]

Alright that makes sense now thankx

Ok can you please have a look at S10 P21 Q4b

So, i use y = (D*lambda) / d to find d. Then I did 1 / d to find the answer. I got it right, but then looked at the MS and saw they used d*sin(theta) = n*lambda to find d. Then they did 1 / d to get the answer.

What im confused about is that when are we meant to know what formula to use where?

The formula you used can only be used when n = 1 that is only one wavelength is involved. It is normally used to find the fringe width which is the distance between the centres of two consecutive bright fringes or that between two dark fringes.

The formula you used proved to be good here since was actually 1. But if that was not the case you would not have been able to find the correct answer. But it's upto you to choose the method you find easier and more appropriate. The essential thing is that you get the required answer.

[$!$RatJumper$!$]

The formula you used can only be used when n = 1 that is only one wavelength is involved. It is normally used to find the fringe width which is the distance between the centres of two consecutive bright fringes or that between two dark fringes.

The formula you used proved to be good here since was actually 1. But if that was not the case you would not have been able to find the correct answer. But it's upto you to choose the method you find easier and more appropriate. The essential thing is that you get the required answer.

I see. I just called my friend now and he says that d*sin(theta) = n*lambda can only be used for diffraction gratings and y = (D*lambda) / d
can only be used for double slit experiments. Is this true?

[TJ-56]

I see. I just called my friend now and he says that d*sin(theta) = n*lambda can only be used for diffraction gratings and y = (D*lambda) / d
can only be used for double slit experiments. Is this true?

Yup. That is correct.

[$!$RatJumper$!$]

In d*sin(theta) = n*lambda, what does the theta actually stand for?

Its really confusing me because for example,

In W03 Q4(b)(ii), they ask for the angle where the first image is seen. It also says the incident light is at right angles. 

But in W06 Q4(b)(i), they ask for the numbers where is it visible so why do we suddenly use a value of 90 for theta in the equation: d*sin(theta) = n*lambda. Wouldnt we need to use the angle like used in W03?

[TJ-56]

In d*sin(theta) = n*lambda, what does the theta actually stand for?

Its really confusing me because for example,

In W03 Q4(b)(ii), they ask for the angle where the first image is seen. It also says the incident light is at right angles. 

But in W06 Q4(b)(i), they ask for the numbers where is it visible so why do we suddenly use a value of 90 for theta in the equation: d*sin(theta) = n*lambda. Wouldnt we need to use the angle like used in W03?

To find the maximum number n of the visible image; we use the eequation you stated,
which is n*lambda=d* sin (theta)
Arranged with n the subject: n = [d* sin (theta)]/ lambda
So to find the max value of n, we need the max value of sin theta, that is when theta is 90 degrees.

[thecandydoll]

May/June 2009 Variant 2 Q2C.
How is the other V=  4.2?
and for  deltap/deltat (how do they find T)
there is one more question similar in 2002/oct/nov how to find the time? when finding force acting on speher?

[TJ-56]

May/June 2009 Variant 2 Q2C.
How is the other V=  4.2?
and for  deltap/deltat (how do they find T)
there is one more question similar in 2002/oct/nov how to find the time? when finding force acting on speher?

for the 2002 paper
see attachment for the required time

for the 2009 paper
Reading carefully, theyre asking about only the first 3.5 seconds, so following 3.5 s gives a velocity of 4.2 m/s

Hope that helped

[$!$RatJumper$!$]

To find the maximum number n of the visible image; we use the eequation you stated,
which is n*lambda=d* sin (theta)
Arranged with n the subject: n = [d* sin (theta)]/ lambda
So to find the max value of n, we need the max value of sin theta, that is when theta is 90 degrees.

Thank you

[TJ-56]

Thank you

Don't mention it.

[TJ-56]

Someone please help with the following, kinda silly but i just want to make sure my answers are correct.

W01 Q3 (c) (i) The correct directions please. ( no need for explanation.)
W09 Q7 P21 (a) Can you post how to correctly point out the angle of deviation.
S10 P21 Q5 (b) (ii) 2. (decreasing)

Thanx for any help!

[Deadly_king]

Someone please help with the following, kinda silly but i just want to make sure my answers are correct.

W01 Q3 (c) (i) The correct directions please. ( no need for explanation.)
W09 Q7 P21 (a) Can you post how to correctly point out the angle of deviation.
S10 P21 Q5 (b) (ii) 2. (decreasing)

Thanx for any help!

Sorry but I won't have time to deal with the 1st question.

Nov 09 P21
7. (a) After you have drawn the two directions, draw straight horizontal line from the particles A and B respectively. Then label each angle of deviation a and b for the specific particles.

Now you just have to mention that b < a

Jun 10 p21
b)(ii)2. Actually it should have been an arrow showing the decrease from positively charged sphere to the earthed metal plate or from the latter to the negatively charged sphere. Indicate the right direction using an arrow.

[Dania]

Refer to the attachment.
Please explain.

[TJ-56]

Sorry but I won't have time to deal with the 1st question.

Nov 09 P21
7. (a) After you have drawn the two directions, draw straight horizontal line from the particles A and B respectively. Then label each angle of deviation a and b for the specific particles.

Now you just have to mention that b < a

Jun 10 p21
b)(ii)2. Actually it should have been an arrow showing the decrease from positively charged sphere to the earthed metal plate or from the latter to the negatively charged sphere. Indicate the right direction using an arrow.

Thanx dk
 i got another,
but i understand if u wont have the time to answer it
S10 QP23 Q3 C (i)

[Deadly_king]

Thanx dk
 i got another,
but i understand if u wont have the time to answer it
S10 QP23 Q3 C (i)

No problem.

I assume you got the previous answers.

Change in K.E + Useful work done = Total k.E in overcoming friction + Change in P.E
1240 + 600 = Total k.E in overcoming friction + 1170
Total k.E in overcoming friction = 670 J

[TJ-56]

On which formula is the one you stated based on?
thanx