Snell’s Law for surface AB: n1 sin60 = n2 sinr1

For air n1 = 1 and n2 = n for a prism as per the question. So as per Snell’s law this equation becomes:

=> √3/2 = n sin r1

=> Sin r1 = √3/2n ——— eq 1

Similarly applying it to surface AC gives

=> n sin r2 = sin θ ——— eq 2

As seen in equilateral triangle ABC , we have

A+B+C = 180⁰ and also A = B = C = 60⁰ [ABC is an equilateral triangle]

Now we have r1 + B = 90⁰ and also r2 + C = 90⁰ so substituting it gives:

60⁰ + 90⁰-r1 + 90-r2 = 180⁰r1+r2 = 60⁰So now lets use simple trigonometry to play with above equation along with eq 1 & 2 to remove the r1 and r2. As per equation 2, we have

n sin r2 = sin θn sin(60-r1) = sin θn [sin 60 cos r1 – cos60 sin r1] = sin θn [√3/2 cos r1 – ½ sin r1] = sin θn [ √3/2 √(1-sin² r1) – ½ √3/2n] = sin θ [substituting equation 1]n [ √3/2 √(1-3/4n) – ½ √3/2n] = sin θn [√3/4 {√(4n²-3) -1} ] = sin θ —————–eq 3 answer to the real question – For n = √3 the value of θis 60 deg and dθ/dn = m . The value of m is?

taking a derivative dθ/dn of equation 3 and substitute the given values i.e. n = √3 & θ= 60. Let’s do that below, applying the operator “d/dn” on both sides:

d/dn [√3/4 {√(4n²-3) -1} ] = d/dn sinθ =(d/dθ sinθ) dθ/dn = cosθ dθ/dndθ/dn = 1/cos θ x d/dn [√3/4 {√(4n²-3) -1} ]After taking a derivative and putting n = √3 & θ= 60 gives

dθ/dn = 2

thnkuuu so much