(II) A parallel beam of light containing two wavelengths, λ1=465 nm and λ2=652 nm, enters the si

step 1 Okay, so we're doing chapter 32, problem 54. So this says a parallel beam of light contained two

(II) A parallel beam of light containing two wavelengths,...

(II) A parallel beam of light containing two wavelengths, $\lambda_{1}=465 \mathrm{nm}$ and $\lambda_{2}=652 \mathrm{nm},$ enters the silicate flint glass of an equilateral prism as shown in Fig. $54 .$ At what angle does each beam leave the prism (give angle with normal to the face)? See Fig. $28 .$

Key Concepts

Snell's Law

This fundamental law of optics relates the refraction angles at the interface between two media with different refractive indices. It allows us to calculate the bend in the light path when a ray passes from one medium to another by using the relationship n? sin(??) = n? sin(??), where n represents the refractive index and ? represents the angle relative to the normal. This principle is essential when determining the paths of light beams through the prism.

Dispersion

Dispersion refers to the dependence of a material's refractive index on the wavelength (or color) of light. In materials like glass, different wavelengths travel at different speeds and are refracted at different angles. This concept explains why a prism separates a beam of mixed colors into its constituent colors by refracting each wavelength differently.

Geometric Optics

Geometric optics deals with the approximation of light as rays that travel in straight lines, and it forms the basis for analyzing the behavior of light in optical systems. This approach is used to trace the paths of light through optical elements like prisms using reflection and refraction principles.

Prism Geometry

Prism geometry involves understanding the shape and internal angles of a prism, which determine how light is internally reflected and refracted. In an equilateral prism, all faces are identical and the internal angles are fixed, allowing for precise calculations of how light enters and exits the prism. This geometric understanding is critical for determining the exit angles of the refracted light beams.

Transcript

00:01 Okay, so we're doing chapter 32, problem 54.

00:06 So this says a parallel beam of light contained two wavelengths.

00:11 We have wavelengths 1 of 465 nanometers.

00:17 The second wavelengths in the light is 652 nanometers.

00:24 These have respective indexes or fractions in the silicate class of 1 .642.

00:34 For the 465 and 1 .619 for 652 nanometer light.

00:46 So they both enter an equal lateral prism, and we want to figure out what angle does each beam leave at.

00:54 So let's draw our prism here, this being 60 degrees.

01:02 Okay, so if we enter in here at angle theta a, we will then leave at an angle theta b, which means we'll enter this one at angle theta b.

01:25 C, sorry, we already did b, and then we'll be leaving an angle theta d.

01:35 So we'll figure out what theta d is for each light.

01:41 Okay, so first let's figure out what theta b is for each.

01:46 So we know, i was right, that smells a lot first.

01:51 We know that the in air, sine of theta a equals n -i for whichever light ray we're thinking about, sine theta -b -i.

02:09 Okay.

02:09 So if we solve that for theta b i, this will be the inverse sign of an a over an i times the sign of theta a.

02:23 Okay, so theta b1 is the inverse sign of 1 over 1 .642 sign of what did it say it entered as? oh, it entered up 45 degrees.

02:44 Let's write this over here.

02:45 Theta a equals 45.

02:48 So 45 degrees.

02:53 And that means for the first light, so for 465 nanometers, we have theta b1 of 25 .51 degrees.

03:04 For b2, we'll plug in the same thing with the different index of refraction.

03:08 We get 25 .90 degrees...

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