isaac physics Rays and Geometry rainbows

Original post by y2467

https://isaacphysics.org/questions/maths_ch7_8_q1?stage=a_level#accchi_m_of_n\

can someone help me with part E?
i used
chi_m = 4*arcsin(sin(i)/n)-2i

and subbing in arccos(sqrt(1/3(n^2-1)) for i

i got

chi_m = 4*arcsin(sin(arccos(sqrt(1/3(n^2-1))))/n)-(2*arccos(sqrt(1/3(n^2-1))))

but this is wrong? can someone look over my substitution

To save a bit of work, what parts are relevant from the earlier parts and what did you get (correct)?

Reply 2

OP

Original post by mqb2766

To save a bit of work, what parts are relevant from the earlier parts and what did you get (correct)?

a) chi == 4*beta - 2*alpha
c) sin(alpha_m) == sin(arccos(sqrt((1)/(3) * (n**(2) - 1))))
d) sin(beta_m) == (sin(alpha_m))/(n)

Reply 3

Original post by y2467

a) chi == 4*beta - 2*alpha
c) sin(alpha_m) == sin(arccos(sqrt((1)/(3) * (n**(2) - 1))))
d) sin(beta_m) == (sin(alpha_m))/(n)

Its one of those a bit fiddly ones which isaac seems to go in for. There is nothing particularly wrong with what youve done, but it seems to want it in a different (simpler) format.

c) is correct but not the "right" expresion for e). So try and get rid of the
sin(arccos
part by noting that its simply a side ratio (opp/hyp). So create a triangle which has a cos = sqrt(n^2-1)/sqrt(3) and use pythaogras to get the opp/hyp side ratio. part c) accepts this answer so you can check as well as your original one. Then use this answer for part e)

Reply 4

OP

Original post by mqb2766

Its one of those a bit fiddly ones which isaac seems to go in for. There is nothing particularly wrong with what youve done, but it seems to want it in a different (simpler) format.

c) is correct but not the "right" expresion for e). So try and get rid of the
sin(arccos
part by noting that its simply a side ratio (opp/hyp). So create a triangle which has a cos = sqrt(n^2-1)/sqrt(3) and use pythaogras to get the opp/hyp side ratio. part c) accepts this answer so you can check as well as your original one. Then use this answer for part e)

got it thanks!

also if you wouldnt mind looking over this https://isaacphysics.org/questions/maths_ch7_8_q3?stage=a_level
i'm not sure how to go about doing part F and G

Reply 5

Original post by y2467

got it thanks!

also if you wouldnt mind looking over this https://isaacphysics.org/questions/maths_ch7_8_q3?stage=a_level
i'm not sure how to go about doing part F and G

Im about to hit the sack, so will look at it in the morning if noone else jumps in tonight.

Reply 6

OP

Original post by mqb2766

Im about to hit the sack, so will look at it in the morning if noone else jumps in tonight.

aight

Reply 7

Original post by y2467

aight

Not worked it through, but the diagram at the end of
http://www-spc.phy.cam.ac.uk/files/rainbows.pdf
should help as it should be very similar to the OP. It looks like this is a draft of the question and Im surprised they didnt include it as a hint/in a description.

Reply 8

OP

Original post by mqb2766

Not worked it through, but the diagram at the end of
http://www-spc.phy.cam.ac.uk/files/rainbows.pdf
should help as it should be very similar to the OP. It looks like this is a draft of the question and Im surprised they didnt include it as a hint/in a description.

ive managed to work out F, but im a little confused for G. what value of n am i supposed to use? do i just sub it in to the equation i got in F and then calculate the difference in angles between the primary and secondary rainbow?

Reply 9

Original post by y2467

ive managed to work out F, but im a little confused for G. what value of n am i supposed to use? do i just sub it in to the equation i got in F and then calculate the difference in angles between the primary and secondary rainbow?

Not really worked it through, but the two angles you got for the two rainbows should be relative to the anti solar point
http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/rbowpri.html
and you want to find the angle the ray from the sun makes with the horizontal.

*** there may be a much simpler way to do this, but ***
So you should be able to estimate the two vertical heights (photo units) above the horizon for each red bands (for example), h1 and h2, and these should be the same horizontal distance, d, from the observer. These correspond to angles alpha and beta with the solar ray, which makes an angle gamma with the horizontal. Then simple trig gives
tan(alpha-gamma) = h1/d
tan(beta-gamma) = h2/d
so rearrange both for d, equate, then expand in terms of tan(alpha), tan(beta) and tan(gamma) using the tan addition/subtraction identity. As youd know everything but tan(gamma) it would give a quadratic in tan(gamma) that can be solved.

As I said at the start, its not necessarily the best way to do it or .... but it is a way and you have to get the fact that your two angles alpha and beta are measured from a common ray somehow.

Edit 1 - an alternative way would be to use the fact that the full rainbow is a circle and the center is where the solar point would be. So construct a couple tangents on the horizon and hence perpendicular chords (or choose two chords and construct their perpendicular bisectors) which would intersect at the centre. Then its just estimating the horizon chord and a bit of trig to get the angle. Probably the simplest, but not the way they want you to do it though.

Edit 2 - a bit of "intelligent" guessing (roughly based on edit 1) and I got it correct so if youve done something 1/2 right but are getting frustrated, just let me know. Its not a great question.

(edited 12 months ago)

Reply 10

WheatenZeus

1

Original post by username5843156

ive managed to work out F, but im a little confused for G. what value of n am i supposed to use? do i just sub it in to the equation i got in F and then calculate the difference in angles between the primary and secondary rainbow?

What did you do to get F and G

Reply 11

Original post by WheatenZeus

What did you do to get F and G

What did you try? There is a reasonable amount of info above.

Reply 12