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# 2002 - multichoice

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Posted 13 May 2006 - 09:45 PM

Questions, 15,17,18

Not sure how to do em.

Any help wud be appreciated.

Thanks!

### #2Michael

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Posted 13 May 2006 - 10:15 PM

15. The waves meet at a second order minimum and the general rule for a minimum is n + 0.5 lambda. As the wavelength is 4 the answer is 1.5 * 4 = 6cm. The first minimum is 0.5 therefore the second minimum = 1.5 Answer = E

16. frequency will remain a constant(this is a learning outcome). n=lamba in air / lambda in glass therefore the answer is 4x10^-7 . Then use c = f x lamda to find the speed. Answer = A

17. Using the diagram the energy of the photon is 1.1 x 10^-19J. If you do 10/1.1x10^-19 you get the answer = E

Hope thats helpful, if you want further explantion post your queries and ill try help further.

Mike

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Posted 14 May 2006 - 05:00 PM

Can somebody explain 3, 13, and 18 aswell please?

Thanks.

### #4Michael

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Posted 14 May 2006 - 05:13 PM

3. 180N acting upwards. The component of weight(mg) acting downwards. Overall unbalanced force of 33N. a=F/m = 33/15 = 2.2 ms-2

13. Saturation is 85% of the supply to the op amp. Therefore the max it can produce is + or - 13V. Since V1 is positive and the op amp is in inverting mode the likely output is -13V.

18. I assume this one works like the inverse square law and the answer would be D

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Posted 16 May 2006 - 01:49 PM

QUOTE(Dhesi @ May 13 2006, 11:15 PM)

15. The waves meet at a second order minimum and the general rule for a minimum is n + 0.5 lambda. As the wavelength is 4 the answer is 1.5 * 4 = 6cm. The first minimum is 0.5 therefore the second minimum = 1.5 Answer = E

16. frequency will remain a constant(this is a learning outcome). n=lamba in air / lambda in glass therefore the answer is 4x10^-7 . Then use c = f x lamda to find the speed. Answer = A

17. Using the diagram the energy of the photon is 1.1 x 10^-19J. If you do 10/1.1x10^-19 you get the answer = E

Hope thats helpful, if you want further explantion post your queries and ill try help further.

Mike

thanks for the help btw, but i still really dont understand 15? real confused...

thanks

### #6Pammy

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Posted 16 May 2006 - 01:59 PM

Thought I would have a go at explaining Q15

path difference= n x wavelength
where n= order number of the maximum/minimum

In this case it is the second order minimum therefore n=1.5 (the first order minimum would be 0.5)

We have been told the wavelength is 4cm= 0.04m

P.d = 1.5 x 0.04
= 0.06
= 6cm

Is that any clearer ?? Hope so
*~* Pam xxx *~*
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### #7*Kiran*

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Posted 16 May 2006 - 05:34 PM

I have been struggling for a while.

### #8Michael

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Posted 16 May 2006 - 05:39 PM

Vo=(V2-V1)Rf/R1 Rf/R1 is the gain and from the original circuit we can find that the gain is 50

The gain remains the same and we use it in the formula:
Vo=(1.2-1.0) x 50 = 10V Answer E

### #9*Kiran*

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Posted 16 May 2006 - 08:19 PM

Thanks Very Muckle!!

much appreciated

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Posted 16 May 2006 - 09:00 PM

QUOTE(Pammy @ May 16 2006, 02:59 PM)

Thought I would have a go at explaining Q15

path difference= n x wavelength
where n= order number of the maximum/minimum

In this case it is the second order minimum therefore n=1.5 (the first order minimum would be 0.5)

We have been told the wavelength is 4cm= 0.04m

P.d = 1.5 x 0.04
= 0.06
= 6cm

Is that any clearer ?? Hope so
*~* Pam xxx *~*

But shouldn't n=2.5 because there are two minimums above the central axis (so n = (2+0.5))

??

### #11mujb

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Posted 16 May 2006 - 09:03 PM

QUOTE(Dhesi @ May 16 2006, 06:39 PM)

Vo=(V2-V1)Rf/R1 Rf/R1 is the gain and from the original circuit we can find that the gain is 50

The gain remains the same and we use it in the formula:
Vo=(1.2-1.0) x 50 = 10V Answer E

i dont understand how you get the gain to be 50, and what original circuit are you talking about??

plz help

### #12Michael

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Posted 16 May 2006 - 09:06 PM

Rf/R1=the gain. So if you use the values they original provide you with you can find the gain. The gain will remain constant in the circuit but they increase the two input voltages which are then amplified by the gain.

### #13mujb

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Posted 16 May 2006 - 09:09 PM

QUOTE(Dhesi @ May 16 2006, 10:06 PM)

Rf/R1=the gain. So if you use the values they original provide you with you can find the gain. The gain will remain constant in the circuit but they increase the two input voltages which are then amplified by the gain.

hi sorry for being annoying but what resistor values are you using. the values provide in q13, because in question 14 it only mentions rf=r3 and r1=r2.
help much appreciated

### #14mujb

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Posted 16 May 2006 - 09:21 PM

hi, its ok i figured it , thanks

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Posted 16 May 2006 - 09:23 PM

Just a quick link to my other post incase anybody missed it.

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Posted 16 May 2006 - 09:44 PM

QUOTE(Betinho @ May 16 2006, 10:23 PM)

Just a quick link to my other post incase anybody missed it.
1.5 is correct, it wouldn't be 2.5. This is because it's the second order minimum, but remember that the maximum in the middle is the zero order maximum so at this point there is zero wavelengths difference between the paths of each wave. Now, for the first minimum, there is going to be 0.5 wavelengths difference between the paths, and for the second minimum there is going to be 1.5 etc.
Lot's of people I know have gotten confused with this, I always tell them to think what order minimum it is (i.e. the second one) and subtract a half from this number to get the number of wavelengths difference. It's probably better if you understand why it is the case, but if not then you can still get the question perfectly correct by doing it this way.
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