Finding the minimum angle of deviation for an equilateral prism

[thezac11]

Homework Statement

Use the formula: n=(sin(A+D)/2)/(sin(A)/2) to find D for an equilateral prism of index of refraction n=1.50

(where A=apex angle for a prism=60 degrees in this case, and D=minimum angle of deviation)

Homework Equations

Snell's Law: (n1)sin(i)=(n2)sin(r) , but i don't think this equation is needed for this problem.

The Attempt at a Solution

This is what I've got, but I know it is not completely correct. Any help would be greatly appreciated:

1.5=(sin(60+D)/2)/(sin(60)/2)

----> cancel the 2's and imput sin(60)cos(D)+cos(60)sin(D) for sin(60+D)

1.5=(sin(60)cos(D)+cos(60)sin(D))/(sin(60))

----> multiply numerator and denominator by cos(D)

1.5=(sin(60)cos(D)+cos(60)sin(D)cos(D))/(sin(60)cos(D))

----> cancel sin(60)cos(D) from numerator and denominator

1.5=cos(60)sin(D)cos(D)

1.5=(0.5)sin(D)cos(D)

3=sin(D)cos(D)

3=(0.5)(2sin(D)cos(D))

3=(0.5)sin(2D)

6=sin(2D)

?

 

[supratim1]

hey!

you cannot cancel the 2! in your second step.

its the angle (A+D)/2 and A/2 whose sine we are taking.

 

[thezac11]

so i didn't cancel the 2's and i came up with:

1.5=((cos(60)sin(D)cos(D))/2)/(1/2)

1.5=((1/2sin(D)cos(D))/2)x2

0.75=(1/2sin(D)cos(D))/2

1.5=(1/2sin(D)cos(D))

1.5=1/2(1/2(2sin(D)cos(D)))

3=1/2(1/2sin(2D))

6=1/2sin(2D)

12=sin(2D)

?

 

[supratim1]

you again did a mistake. consider angle A/2 = x and D/2 as y, and then proceed. you will grasp the idea.