physics 11 - weeblylight rays when sunlight enters an equilateral crown glass prism at an angle of...

Physics 11Unit 8 – Geometric Optics

Part 3

(d) Dispersion

• When a monochromatic light enters into a glass prism, it is refracted toward the surface normal because the refractive index of glass is higher than that of air. When it leaves the prism, it is bent away from the normal further. The net effect is that the light gets deflecteddownward by the prism.

Unit 8 - Geometric Optics (Part 3) 2

• However, when sunlight travels through a prism, it is observed that the different color lights are separated, and are refracted at different angles. As a whole, a white light is split into different color lights by the prism. This phenomenon is called dispersion.

• The effect of dispersion is due to the dependence of the refractive index on the wavelength (or frequency) of the incident light. In general, for transparent substances such as glass, the higher the frequency of the incident ray (or the smaller the wavelength), the larger the refractive index.


• Example: A ray of sunlight is passing from diamond into crown glass; the angle of incidence is 35°. The indices of refraction for the blue and red components of the ray are: blue (ndiamond = 2.444, nglass = 1.531) and red (ndiamond = 2.410, nglass = 1.520). Determine the angle between the refracted blue and red rays in the crown glass.

• Recall the Snell’s law: 𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2.

• The angle of refraction for red light is:

• The angle of refraction for blue light is:

Unit 8 - Geometric Optics (Part 3)

𝜃𝑟 = sin−1𝑛𝑑𝑛𝑔

sin 𝜃𝑑 = sin−12.410

1.520sin 35° = 65.4°

𝜃𝑏 = sin−1𝑛𝑑𝑛𝑔

sin 𝜃𝑑 = sin−12.444

1.531sin 35° = 66.3°

4

• Hence the angle of deviation between the red and blue light in crown glass is:


∆𝜃 = 𝜃𝑏 − 𝜃𝑟 = 66.3° − 65.4° = 0.9°

5

• Practice: What is the angle of deviation of the emitting red and violet light rays when sunlight enters an equilateral crown glass prism at an angle of 50°?


• An illustrative example of dispersion of light is rainbow which is formed by the refraction of light by water droplets.


• When sunlight enters into a spherical rain droplet, light of each color is refracted by the amount depending on the corresponding refractive index in water. After being reflected at the back of the droplet, different colors are refracted again and re-enter into air.

• Since the light rays of different colors are refracted to different extent, they leave the water droplet in different directions. Hence, an observer only sees one color of light coming from a droplet at a given direction.


• Despite that, we are still able to see the full color spectrum of a rainbow because each color of a rainbow comes from different droplets at different angles of elevation.


(e) Interference

(i) The principle of Superposition

• When two sound waves meet one another at a given point at the same time, the loudness of the resulting sound is altered because of the principle of superposition.

• Likewise, when two light rays, which are electromagnetic waves, pass the same position at the same time, they do interfere causing the change of the brightness, which is related to the intensity, of the resulting light.

• The two light waves may interfere constructively or destructivelydepending on whether they are in-phase or out-of-phase.


• Constructive interference: It occurs when the distances between the point P and the two sources, respectively, differ by a integral multiple of the wavelength λ.


• Destructive interference: It occurs when the distances between the point P and the two sources, respectively, differ by a half-integral multiple of the wavelength λ.


(ii) Young double-slit experiment

• This is an experiment originally performed by Thomas Young in 1901, with the purpose of demonstrating the wave nature of light. The major hypothesis behind the experiment is that if light is wave, two light rays produced by coherent light sources will interfere each other, resulting in a pattern consisting of alternating bright and dark fringes.

• A simple setup for the Young’s experiment:


• The following diagrams explain how bright and dark fringes are formed.

• The key is the difference between the two path lengths ℓ1 and ℓ2.

• If ℓ2 − ℓ1 = 𝑛λ, then a bright fringe is formed. (Constructive)

• If ℓ2 − ℓ1 =𝑛

2λ, then a dark fringe is formed. (Destructive)


• The fringes in the interference pattern have different level of intensity; the central fringe is the brightest, and the fringes at either side have the intensities decreasing symmetrically with increasing distance from the central fringe.


• The positions of the fringes can be calculated with an aid of the following diagrams:

• If the screen is very far from the double slit, the two light rays ℓ1 and ℓ2 can be considered parallel. The path difference ∆ℓ is therefore dependent upon the slit separation 𝑑 and the angle 𝜃:


• Bright fringe (i.e., constructive interference) occurs when ∆ℓ equals 𝑚λ; hence

• Similarly, dark fringe (i.e., destructive interference) occurs when ∆ℓ

equals 𝑚 +1

2λ; hence


∆ℓ = 𝑑 sin 𝜃

sin 𝜃 = 𝑚λ

𝑑𝑚 = 0, 1, 2, …

sin 𝜃 = 𝑚 +1

2

λ

𝑑𝑚 = 0, 1, 2, …

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• Example: Red light (λ = 664 nm) is used in Young’s experiment with the slits separated by a distance of 1.20 x 10-4 m. The screen is located at a distance of 2.75 m from the slits. Find the distance on the screen between the central fringe and the third order bright fringe.


• Note that the separation between the slits and the screen is much larger than the distance between the two slits. Therefore, the approximation ∆ℓ = 𝑑 sin 𝜃 is valid. Hence, for bright fringes,

• For 𝑚 = 3, λ = 660 nm, and 𝑑 = 1.20 × 10−4m:

• Therefore


sin 𝜃 = 𝑚λ

𝑑

𝜃 = sin−1 3660 × 10−9

1.20 × 10−4= 0.95°

𝑦 = 𝐿 tan 𝜃 = 2.75 tan 0.95° = 0.045 m

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(iii) Diffraction grating

• The interference patterns of bright and dark fringes occur when monochromatic light passes through a double slit. Such patterns can also be formed when light passes through more than two slits. The arrangement consisting of a large number of parallel, closely spaced slits is called a diffraction grating.


• The principle of the formation of grating patterns is identical to that for the double-slit experiment.


• Different bright fringes are formed corresponding to the interference of two light rays whose path length difference is a multiple of λ.


sin 𝜃 = 𝑚λ

𝑑𝑚 = 0, 1, 2, …

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• Example: A mixture of violet light (λ = 410 nm) and red light (λ = 660 nm) falls on a grating that contains 1.0 × 104 lines/cm. For each wavelength, find the angle 𝜃 that locates the first order maxima.

• The slit separation is Τ0.0110000 = 1.0 × 10−6 m.

• For red light:

• For violet light:

• The first-order red and violet fringes are seen separately on the screen.Unit 8 - Geometric Optics (Part 3)

𝜃𝑟 = sin−1 3660 × 10−9

1.0 × 10−6= 41°

𝜃𝑣 = sin−1 3410 × 10−9

1.0 × 10−6= 24°

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• This phenomenon is manifested in particular when sunlight passes through a grating. The central fringe is white since all color lights converge there. Each of the other fringes is made of a full spectrumof colors from violet to red with increasing 𝜃.

• Usually adjacent spectra corresponding to high-order fringes overlapbecause of their large spans of 𝜃.


(iv) Other interference patterns

• The followings are some interesting and illustrative examples of interference we may encounter in daily life.


physics 11 - weeblylight rays when sunlight enters an equilateral crown glass prism at an angle of...

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