Ocr Physics B Paper 2 June 8th 2018

What was the last 6 marker again?

Reply 7

kathschof1

5

Original post by Lavalamp1999

What was the last 6 marker again?

About the non-reflective coating on the solar cells. It gave you their refractive indexes and the wavelength of light going in. I had no idea what to say lol.

Reply 8

Original post by kathschof1

About the non-reflective coating on the solar cells. It gave you their refractive indexes and the wavelength of light going in. I had no idea what to say lol.

Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.

Reply 9

kathschof1

5

Original post by Lavalamp1999

Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.

Oh wow! I would never have thought of that. Thanks so much for your help!

Reply 10

Vanilla Cupcake

15

what we thinking in terms of grade boundaries? found it a lot harder than last year's paper 2 so im really hoping for it to be lower lol. there were a lot of calculations i was surprised but that one about the bananas over 20 years was tragic. hoping they're generous with method marks

Reply 11

Original post by Lavalamp1999

Oh yeah. The wavelength decreases by the same proportion as the refractive index increases. So by the time the phasor passes through the coating and bounces back it has rotated by pi radians, destructively interfering with the top surface > less reflecting.

Also the fact that light slows down when changing to a higher index medium so that the angle of incidents are much lower so less total internal reflection. You could support this by doing a simple calculation of the critical value increases as it switches mediums.

Reply 12

Original post by poseidon820

How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.

What did everyone

Original post by poseidon820

How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.

What did everyone get for the efficiency on the the last equation

Reply 13

Tom19800

9

Original post by Dahen2323

What did everyone

What did everyone get for the efficiency on the the last equation

I just ran out of time half way through the question, so annoyed. I was going to do the intensity is proportional to 1/r^2, so 1/5.2^2 = the intensity at 5.2AU. Then I'd find the efficiency using the difference in power from the value i'd calculated? I might be remembering the question incorrectly though.

Reply 14

Original post by Dahen2323

What did everyone

What did everyone get for the efficiency on the the last equation

16%

Reply 15

Vanilla Cupcake

15

Original post by Lavalamp1999

16%

good to know i got one thing right

Reply 16

Original post by Tom19800

I just ran out of time half way through the question, so annoyed. I was going to do the intensity is proportional to 1/r^2, so 1/5.2^2 = the intensity at 5.2AU. Then I'd find the efficiency using the difference in power from the value i'd calculated? I might be remembering the question incorrectly though.

Yea that’s what I done

Reply 17

Original post by poseidon820

How did everyone find it? I found it much better than Monday's paper but that last 6 marker was awful.

What about the question about does the solar pulse have enough energy to move from 1500 m to 2300m or whatever it was. Could you use mgh for gravitational potential energy change as you didn’t have the mass of earth and the radius ?

Reply 18