Question

A light beam strikes a piece of glass at a 60◦incident angle. The beam contains two wavelengths,450 nm and 700 nm, for which the index of refraction of the glass is 1.4820 and 1.4742, respectively.What is the angle between the two refracted beams?

Answer 1

Answer:

α = 0.22°

Step-by-step explanation:

To find the angle between the two refracted beams we need to use Snell's Law:

$n_{1}sin(\alpha_{1}) = n_{2}sin(\alpha_{2})$  

Where:

n₁: is the refractive index of the incident beam (air) = 1

n₂: is the refractive index of the refractive beam (glass)

α₁: is the angle of the incident beam = 60°

α₂: is the angle of the refractive beam  

Hence, for the beam with λ = 450 nm and n₂ = 1.4820, the refractive angle (α₂) is:

$sin(\alpha_{2}) = \frac{n_{1}}{n_{2}}*sin(\alpha_{1}) = \frac{1}{1.4820}sin(60) = 0.5844$                  

$\alpha_{2} = 35.76 ^{\circ}$

Similarly, the refractive angle for the beam with λ = 700 nm and n₂ = 1.4742 is:

$sin(\alpha_{2}) = \frac{n_{1}}{n_{2}}*sin(\alpha_{1}) = \frac{1}{1.4742}sin(60) = 0.5875$                  

$\alpha_{2} = 35.98 ^{\circ}$

Finally, the angle between the two refracted beams is:

$\alpha_{2}_{(\lambda=700nm)} - \alpha_{2}_{(\lambda =450nm)} = 35.98 ^{\circ} - 35.76 ^{\circ} = 0.22 ^{\circ}$

I hope it helps you!