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  θn with the normal (see the figure).  For n=3  the value of  θ is  60° and  dθdn=m.  The value of  m is

We know, r1+r2=A=60∘  (equilateral prism) Snell's Law at P, (1) sin⁡60∘=(n)sin⁡r1⇒32=(n)sin⁡60∘−r2⇒32=(n)sin⁡60∘cos⁡r2−cos⁡60∘sin⁡r2⇒32=(n)321−sin2⁡r2−12sin⁡r2−−−−−(1) Snell's Law at Q , nsin⁡r2=(1)sin⁡θ⇒sin⁡r2=sin⁡θn−−−−−−(2)Put⁡(2)in⁡(1)⇒32=(n)321−sin2⁡θn2−12sin⁡θnAfter squaring both sides,⇒4sin2⁡θ+23sin⁡θ−3n2+3=0 Differentiating wrt n , ⇒4(2sin⁡θ)cos⁡θdθdn+23cos⁡θdθdn−6n=0 Put θ=60∘ and n=3⇒dθdn=2 So ,m=2