11i forgot to mention the refractive index is 1.44....
The rays are parallel to each other.
What angle will the emerging rays will make with each other?
please give me the steps.
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15 Answers
11is the answer 20o ...confirm then i'll post the solution....
11
Please tell where am I going wrong....
This is how I solved...
since incedent angle is 30o (< 44o (ic)) the refracted angle should be 20o....
Pls point out where am I going wrong...
11Why are u taking the incident angle as 30°?the answer given is 2[arcsine(0.72)-30°],but i can't figure why.
it means 32° is right...ya..thanks...
11Considering just the one half of the prism...
angle of incidence is the angle b/w the incident ray and the normal
so the angle of incidence is 30o
infact Arka has also taken the incident angle as 30o...but how are we getting the refracted angle as 46o..?????
11yups
and
sin 30sin r = 1.44
sin r = sin 301.44 = 12 x 1.44 = 12.88 = 0.3472
r = sin-1 (0.3472) = 20.31o ≈ 20o
so how 46o????
1sandipan,a little mistake on ur part..
refractive index is given as 1.44.
which is refractive index of glass with respect to air(aμg)
when u are considering the refraction of the rays out of the prism,ie from glass to air,sin 30/sin r=gμa
which is refractive index of air with respect to glass which will be 1/1.44.
from here we get r as 46°.
also from denser to rarer medium =>ray goes away from the normal(which isnt the case in ur solution).