Question

Q.A monochromatic beam of light is incident at 60° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For n = 3 the value of θ is 60° and d θ d n = m . The value of m is

Answer 1

Solution:  Let me explain the solutions with set of concepts used to solve this question, except concept 1 other concepts are purely based on mathematics:

Concept 1: 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

Concept 2: Here we need to use a property of the triangle learned in class 9th!

So, we have two equations above with 4 variables. We need only some relationship between θ and “n” to solve this problem. Because if we find an equation involving these two variables then it’s easy to take a derivative and solve. Let’s see how a property of equilateral triangle will help here. 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 = 60⁰

So 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

Concept 3: So, we have now equation 3 where only two variables are present i.e. n and θ. Next job is to find the answer to the real question – For n = √3 the value of θ is 60 deg and dθ/dn = m . The value of m is?

Now its simple stuff, what we need is just 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