Light consisting of a mixture of red and blue light enters a 40°, 70°, 70° prism along a line parallel to the side opposite the 40° vertex. the index of refractio n of the prism material for blue light is 1.530, and for red light it is 1.525. what is the angle between the two emergin g beams of light?

This is actually a simple problem of trigonometry

first take a look at the picture
you need to understand that the incident beam of light meets the left side of the prism at a 70° angle wich makes angle v = 20°
then that left normal and right normal make an 140° angle (those normals with the upper corner of the prism form a 4-side polygon with the sum of all angles 360°, two of its angles are 90° and the upper one is 40°, therefore the 4th angle is 360-90-90-40=140)

ok, now comes the only physics part of the problem
you only need to know the refractive index formula
n(air)*sinv = n(prism for blue)*sinb
n(air)*sinv = n(prism for red)*sinr

so we have 
sinv = 1.530*sinb => sinb= sin20°/1.530 = 0.342/1.530 = 0.223 => b=12.92°
sinv = 1.525*sinr => sinr = sin20°/1.525 = 0.342/1.525 = 0.224 => r=12.96°

now, we know that the sum of a triangle angles is 180° so r+r1+140=180
therefore r1=27.04
same way we get b1 = 24.08

then again we apply the refractive index formula
and we get
sinr2 = 1.525*sin27.04° = 1.525*0.455 = 0.693 => r2 = 43.86°
sinb2 = 1.530*sin24.08° = 1.530*0.408 = 0.624 => b2= 38.62°

now the angle between blue and red rays is r2-b2 = 5.24°

Ithink the answer is a