bynary Posted May 8, 2016 Report Share Posted May 8, 2016 2 minutes ago, Polybos said: Your calculation is good. The problem is that it was a trick question. You assume it was unpolarized light that hit the polarizer which is what the IB usually does. However this time the light was horizontally polarized initially. If you want you can think of it as having three polarizers. One horizontal, one that rotates and one vertical. Then you can use your formula and see that cos 90 = 0. Nowhere do I assume that it is unpolarized. The axis of polarization goes from horizontal to at an angle (45° in my example) which is perfectly valid. Then from a 45° angle it meets a vertical polarizer and it gets polarized vertically which is also valid. 2 Reply Link to post Share on other sites More sharing options...
Polybos Posted May 8, 2016 Report Share Posted May 8, 2016 2 minutes ago, indimpi said: http://physics.stackexchange.com/questions/61918/three-polarizers-45-apart This illustrates the problem from the exam, light does in fact come through. Guess I'm wrong. Sorry guys 2 Reply Link to post Share on other sites More sharing options...
bynary Posted May 8, 2016 Report Share Posted May 8, 2016 Just now, Polybos said: Guess I'm wrong. Sorry guys massive props to admitting... most people would have just left Reply Link to post Share on other sites More sharing options...
Polybos Posted May 8, 2016 Report Share Posted May 8, 2016 9 minutes ago, bynary said: massive props to admitting... most people would have just left Guess I know why these people are doing their petition. This was nowhere in the book we have. Looking back our teacher had actually thought us this though so props to him, Impressed someone got it right though. Reply Link to post Share on other sites More sharing options...
ibstudent081099 Posted May 8, 2016 Report Share Posted May 8, 2016 1 hour ago, Polybos said: When you divide the diameter by two the absolute uncertainty is also divided by two. So percentage uncertainty stays the same. Therefore when you cube it the percentage uncertainty is the same anyway. Just think of it as doing d/2 in the formula for the sphere. 2 is just a constant so it doesn't matter for percentage uncertainty as it does not have an uncertainty itself. This is quite important for paper 3 so I would advise you to learn it. For the polarizer question the light was already horizontally polarized, so when it went through the vertical polarizer nothing would go through regardless of what the first polarizer was. A bit of a trick question, but oh well. Is the percentage uncertainty the fractional uncertainty multiplied by 100? Reply Link to post Share on other sites More sharing options...
Polybos Posted May 8, 2016 Report Share Posted May 8, 2016 Just now, ibstudent081099 said: Is the percentage uncertainty the fractional uncertainty multiplied by 100? Yeah. Reply Link to post Share on other sites More sharing options...
ibstudent081099 Posted May 8, 2016 Report Share Posted May 8, 2016 14 minutes ago, Polybos said: Yeah. Because I put the answer with pi. In the data booklet the formula for the uncertainty shows that you have to multiply by the volume. As a consequence I got pi in the value. Could anyone tell me if this is wrong because I will need it for tomorrow. Thanks guys Reply Link to post Share on other sites More sharing options...
Polybos Posted May 8, 2016 Report Share Posted May 8, 2016 Pi does not have any uncertainty as it is a constant and should not be included in fractional uncertainty or percentage uncertainty. If you multiply by pi both the value itself and its absolute uncertainty gets multiplied by pi, which is why fractional does not change. Reply Link to post Share on other sites More sharing options...
ibstudent081099 Posted May 8, 2016 Report Share Posted May 8, 2016 Just now, Polybos said: Pi does not have any uncertainty as it is a constant and should not be included in fractional uncertainty or percentage uncertainty. If you multiply by pi both the value itself and its absolute uncertainty gets multiplied by pi, which is why fractional does not change. I multiplied by pi because when calculating the uncertainty the formula in the data booklet is uncertainty y:y pi is included in the value of y. when calculating the uncertainty I thought we needed to multiply the by value of y. I hope what I meant is clear and not confusing lol Reply Link to post Share on other sites More sharing options...
ibstudent081099 Posted May 8, 2016 Report Share Posted May 8, 2016 9 minutes ago, Polybos said: Pi does not have any uncertainty as it is a constant and should not be included in fractional uncertainty or percentage uncertainty. If you multiply by pi both the value itself and its absolute uncertainty gets multiplied by pi, which is why fractional does not change. This is what I meant Reply Link to post Share on other sites More sharing options...
Meep202 Posted May 10, 2016 Report Share Posted May 10, 2016 Did anyone know how to answer the question in paper 2 about calculating the temperature of earth? It was part of the albedo and energy from the sun question. Reply Link to post Share on other sites More sharing options...
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