Diffraction of Light. Huygens-Fresnel Principle

a. Determine the angle of inclination of the wedge .

Lesson 2.8.

Lesson 2.9.

Two parallel monochromatic rays of light shine on the upper surface of a thin plate with a surface of

2

o

thickness d constant, refractive index n 1 with angle of incidence i 1 = 45 . Above the thin plate is

air ( n o =1 ), below is glass ( n 2 =1.5 ) thickness d=0.49µm , light wavelength  0.6  m .

b. At M is the bright point or the dark point.

Lesson 2.10.

Shine a parallel monochromatic beam of light perpendicular to the flat glass plate of the device to produce Newton's circular interference fringes. Observe that the radii of two consecutive dark fringes are 4.38mm and 4.73mm respectively . The radius of the spherical cap R=6.4m . The refractive index of the glass of the device is n=1.5 .

a. Explain the formation of Newton's circular interference pattern. Give the order number of the two dark fringes mentioned above and calculate the wavelength of the light.

b. If transparent water ( n1 = 4 /3 ) is filled into the gap between the spherical cap and the flat glass plate of the instrument, how will the interference pattern change?

Lesson 2.11.

An apparatus for Newton's rings consists of a thin glass lens.

A flat-concave crystal ( n 2 = 1.5 ) with a concave surface radius of R = 2m is placed so that the concave surface faces up and is filled with transparent water ( n 1 = 4/3 ). A parallel monochromatic beam of light (wavelength ) is projected onto it. 0.65  m ) perpendicular to the water surface. Determine the thickness of the glass layer at the 10th dark fringe position and calculate the radius of that dark fringe.

Lesson 2.12.

The interference device for the Newtonian ring system consists of a flat horizontal glass cap (with the same refractive index n = 1.5 ). The radius of the cap is R = 8.6m . A parallel, monochromatic beam of light is shined perpendicular to the glass plate. The Newtonian ring interference pattern is observed, the diameter of the fourth dark fringe is 9mm . The center of the pattern is considered to be the zeroth dark fringe. The gap between the cap and the glass plate is air ( n o = 1 ).

a. Calculate the wavelength of light incident on the interference device.

Lesson 2.13.

Two flat glass plates ( n=1.5 ) are tilted to each other at a very small angle α, forming an air wedge between the slits. Shine a monochromatic beam of light

 0.6  m perpendicular to the bottom surface of the air wedge. Observe the interference pattern on the wedge surface and on 1cm long of the wedge surface, there are 11 dark fringes.

a. Calculate the angle of inclination of the wedge.

Lesson 2.14.

Shine a beam of white light onto a soap film (considered a thin plate with variable thickness, refractive index n=1.33 ). Find the smallest thickness of the soap film at which there are yellow reflected rays. Consider two cases:

a. The angle of incidence is 45 o .

b. The incident beam is perpendicular to the soap film.

Given the wavelength of yellow light

Lesson 2.15.

 0.6  m .

Lesson 2.16.

Lesson 2.17.

Given a parallel, chromatic beam of light shining perpendicularly on the plane of a thin air plate between a flat glass plate placed in contact with the curved surface of a plano-convex lens . The radius of the convex surface of the lens is R=15m .

Chapter 3.DIFRACTION OF LIGHT

3.1. LIGHT DIFFRATION PHENOMENON. HUYGENS-FRESNEL PRINCIPLE

3.1.1. Light diffraction phenomenon

In geometrical optics in a homogeneous medium, light travels in a straight line. However, experiments have shown that this is not always true. Consider the following two experiments:

Experiment 1:Use a flashlight S to shine on a circular hole with diameter BD on a thick cardboard P (Figure 3.1). According to the law of rectilinear propagation of light, we can only observe light in the cone SB'D' created by the light rays passing through the edge of the circular hole. However, if we place our eyes

At point M outside and even quite far away this cone still receives light from S .

Figure 3.1

This proves that light propagates into the geometric shadow region, which means it does not obey the law of rectilinear propagation of light.

Experiment 2:Place a thick, circular piece of cardboard O parallel to a light slit. Behind the cardboard, place an observation screen E (Figure 3.2) parallel to O. According to the law of rectilinear propagation, the region AB is covered.

The region hidden by the string must be a geometric shadow region and the region outside must be uniformly illuminated. However, at point M on

Figure 3.2

The axis of symmetry of AB is in the geometric shadow region, we still see light.

and in the vicinity of points A and B we see dark and bright fringes.

This shows that in this case light also does not obey the law of rectilinear propagation.

In both experiments above, the perforated screen BD and the opaque cardboard O are the obstacles that play the role of redistributing the light intensity on the observation screen.

The phenomena observed in those two experiments are examples of light diffraction.

Definition​

obstacle

call

how much

light radiation

how much

far from light

can be explained qualitatively

melon

on Huygens principle.

However, Huygens' principle

does not indicate how the light intensity on the screen placed behind the obstacle will be distributed. To solve this problem Fresnel

Figure 3.3

added the interference principle to Huygens' principle and established the Huygens-Fresnel principle. That is the basic principle of wave optics.

3.1.2. Huygens-Fresnel principle

a. Statement

Lip

point in space

light waves from real sources

both become secondary light sources that emit light waves forward.

cause ear

location of the source​

Figure 3.4. Qualitative explanation of light diffraction phenomenon

Each secondary light source on the circular hole BD has an amplitude and phase of oscillation exactly equal to the amplitude and phase of oscillation caused by light source S at that point. The light oscillation at each point on the screen E will be equal to the sum of the light oscillations caused by the secondary light sources on the circular hole BD at that point . From the expression

thank you expression​​

By Huygens - Fresnel principle one can find

g of the light oscillation at a point M on the screen E ,

but the calculation is quite complicated because it requires integrals . Fresnel gave

an alternative method of calculation called the Fresnel sphere method which we​​​

Determine the light oscillation at M.

We apply the Huygens-Fresnel principle to write the expression for the light oscillation caused at M.

The point light source O can be replaced by a closed luminous surface S surrounding O. According to Huygens' principle, each point dS on surface S

both receive light sent by O so dS

Thus, to find the wave expression at M outside surface S, we do not need to pay attention to O but only need to use the secondary sources dS distributed on surface S.

According to Fresnel's principle, the light oscillation at dS sent by O has the form: