So here for part A. We know that the angle of incidence fate US of B one, uh, at be is a compliment of the critical angle at a and so it's we can say sign of fate. US of B one would be equal to co sign of the critical angle. Or we can say the square root of one divided by ah, the refractive index of medium three, divided by the refractive index of medium to quantity squared. And so, uh, Fada sub B two is gonna be equal to arc sine of and sub two divided by ends up three multiplied by, uh, square root of one minus the refractive index of ah, medium three, divided by ends up to quantity squared, closed parentheses. This is gonna equal arc sine of the square root of and substitute. What about ends up three quantity squared minus one close parentheses and we find out the status of B two will be equal to 35 0.1 degrees. This would be our answer for part A. For part B. We know that end sub one sign of fada would be equal to ends up to sign of fajita sub sea and this would be equal to ends up to times, ends up three divided by ends up to. And so we simply find that here theta is going to be equal to arc sine Uh, and step three divided by and serve one. And we know that this is going to be 49.9 degrees. This would be our answer for part B for part C. We know that here Ah, the angle of incidence a sub a status of a one at a is compliment of the critical angle at B s, we can say sign, Uh, Seita sub a one is equaling co sign of the critical angle fate of subsea, which would equal one, uh rather the square root of one minus and sub three divided by ends up to quantity squared, and we can then solve for the angle of refraction at a angle of incidence. Rather, at a this unequal to arc sine of things, the arc sine of ends up to over ends up three multiplied by the square root of one minus and step three over ends up to quantity squared and this will become status of a one would become arc sine of the square root of and servitude about about ends up three quantity squared minus one and this is equaling again. 35.1 degrees. This would be your answer for park. Si ah, form part D. We know that here ends of one sign of data will equal and sub to sign of status of a one. Ah, this vehicle to ends up two times the square root of one minus ends up three divided by ends up to quantity squared and this is gonna equal to the square root of and sub two squared minus and sub three squared. And then we find Seita for parties, data to be equal to arc sine of the square root of and sub two squared minus sense of three squared, divided by and someone and we find that data is going to be 26.1 degrees. This would be your answer for part D Finally, for rather four part F for party. Rather, we can say that the angle of refraction status of B two at B. This is the fate of the angle of refraction. We can say that fate us A. B two would be equal to arc sine hear of ends up two squared, divided by ants of three, divided by a rather multiplied by the square root of ends up two squared plus and sub three squared, and this is equaling 60.7 degrees. This would be our answer for Part E. And so we can then say that four part F knowing ends and someone sign of theta equals and sub, too. Sign of the Brewster's angle. This would've eagle two ends up to multiplied by ends up three, divided by answer two squared plus and sub three squared. Knowing that here, ah, the angle of incidence fate us of B one at B is the compliment of the Brewster's angle at a and so, uh, we couldn't just substitute for the Brewster's angle, and we find that here Fada would be equal to arc sine of this of and sub two times and sub three. Divided by ants of one times the square root of ends up two squared plus and sub three squared close parentheses. This is equaling 35.3 degrees. This would be our answer for part F. That is the end of the solution. Thank you for watching