Question

(*) A narrow beam of light containing yellow (580 nm) and green (50nm) wavelengths goes from polystyrene to air, striking the surface at a 22.5 incident angle. What is the angle (in degrees) between the colors when they emerge? (b) How far would they have to travel (nm) to be separated by 1.00 mm?

Answer 1

Solution:

Given,

angle of incidence $i=22.5^0$

refractive index of polysterene for yellow light $n_{y}=1.492$

refractive index of polysterene for green light

ng 1.498

Let the angle of refraction for yellow and green be
代
and $r_{g}$ .

(a) Ans: $\Delta \theta=0.027^0$

Using Snell's law,

For Yellow;

$r_{y}=\sin^{-1}\frac{n_{y}\sin i}{n}=\sin^{-1}\frac{1.492\times \sin22.5}{1}=34.817^0$

For Green;

  $r_{g}=\sin^{-1}\frac{n_{g}\sin i}{n}=\sin^{-1}\frac{1.493\times \sin22.5}{1}=34.844^0$

Thus the angle of seperation between the colours when they emerge is given by,

  $\Delta \theta= r_{g}-r_{y}=34.844-34.817=0.027^0$

(b) Ans: 2.12m

The seperation between the rays is 1.0 mm.

$\Delta \theta=0.027^0=4.712\times 10^{-4}rad$

Since $\Delta \theta< <<<0$ , we can take seperation as the arc and the distance rays have to travel as the radii. So,

  $R=\frac{S}{\Delta \theta}=\frac{1.0\times10^{-3}}{4.712\times10^{-4}}=2.12m$