in young's experiment using monochromatic light the fringe pattern shifts

In a Young'S Double Slit Experiment, Using Monochromatic Light, the Fringe Pattern Shifts by a Certain Distance on the Screen When a Mica Sheet of Refractive Index 1.6 - Physics

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In a Young's double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron (1 micron = 10 −6  m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.

Solution Show Solution

Refractive index of the mica sheet,μ = 1.6

Thickness of the plate,

\[t = 1 . 964 \text{ micron }= 1 . 964 \times  {10}^{- 6}   m\]

Let the wavelength of the light used = λ.

Number of fringes shifted is given by

\[n = \frac{\left( \mu - 1 \right)t}{\lambda}\]

So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.

\[\text{Shift} = n   \times   \beta\]

\[ = \frac{\left( \mu - 1 \right)t}{\lambda} \times \frac{\lambda D}{d}\]

\[ = \frac{\left( \mu - 1 \right)t \times D}{d}..........(1)\]

As per the question, when the distance between the screen and the slits is doubled,

i.e.  \[D' = 2D\]

fringe width,

\[\beta = \frac{\lambda D'}{d} = \frac{\lambda 2D}{d}\]

According to the question, fringe shift in first case = fringe width in second case.

\[\text{So, }\frac{\left( \mu - 1 \right)t \times D}{d} = \frac{\lambda2D}{d}\]

\[ \Rightarrow \lambda = \frac{\left( \mu - 1 \right)  t}{2}\]

\[             = \frac{\left( 1 . 6 - 1 \right) \times \left( 1 . 964 \right) \times {10}^{- 6}}{2}\]

\[             = 589 . 2 \times  {10}^{- 9}  = 589 . 2\text{ nm}\]

Hence, the required wavelength of the monochromatic light is 589.2 nm.

Video Tutorials VIEW ALL [3]

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Young's Double-Slit Experiment ( AQA A Level Physics )

Revision note.

Double Slit Interference

• The interference of two coherent wave sources 
• A single wave source passing through a double slit
• The laser light source is placed behind the single slit
• So the light is diffracted, producing two light sources at slits A  and B
• The light from the double slits is then diffracted, producing a diffraction pattern made up of bright and dark fringes on a screen

The typical arrangement of Young's double-slit experiment

Diffraction Pattern

• Constructive interference between light rays forms bright strips, also called fringes , interference fringes or maxima , on the screen
• Destructive interference forms dark strips, also called dark fringes or minima , on the screen
• Each bright fringe is identical and has the same width and intensity
• The fringes are separated by dark narrow bands of destructive interference

The constructive and destructive interference of laser light through a double slit creates bright and dark strips called fringes on a screen placed far away

Interference Pattern

• The Young's double slit interference pattern shows the regions of constructive and destructive interference:
• Each bright fringe is a peak of equal maximum intensity
• Each dark fringe is a a trough or minimum of zero intensity
• The maxima are formed by the constructive interference of light
• The minima are formed by the destructive interference of light

The interference pattern of Young's double-slit diffraction of light

• When two waves interfere, the resultant wave depends on the path difference between the two waves
• This extra distance is the path difference

The path difference between two waves is determined by the number of wavelengths that cover their difference in length

• For constructive interference (or maxima), the difference in wavelengths will be an integer number of whole wavelengths
• For destructive interference (or minima) it will be an integer number of whole wavelengths plus a half wavelength
• There is usually more than one produced
• n is the order of the maxima or minima; which represents the position of the maxima away from the central maximum
• n  = 0 is the central maximum
• n  = 1 represents the first maximum on either side of the central,  n  = 2 the next one along....

Worked example

Determine which orders of maxima are detected at M as the wavelength is increased from 3.5 cm to 12.5 cm.

The path difference is more specifically how much longer, or shorter, one path is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths.

Fringe Spacing Equation

• The spacing between the bright or dark fringes in the diffraction pattern formed on the screen can be calculated using the double slit equation:

Double slit interference equation with w, s and D represented on a diagram

• D  is much bigger than any other dimension, normally several metres long
• s is the separation between the two slits and is often the smallest dimension, normally in mm
• w  is the distance between the fringes on the screen, often in cm. This can be obtained by measuring the distance between consecutive maxima or minima.
• The wavelength of the incident light increases
• The distance between the screen and the slits increases
• The separation between the slits decreases

Dependence of the interference pattern on the separation between the slits. The further apart the slits, the closer together the bright fringes

Calculate the separation of the two slits.

Since w , s and D are all distances, it's easy to mix up which they refer to. Labelling the double-slit diagram as shown in the notes above will help to remember the order i.e. w and s in the numerator and D underneath in the denominator.

Interference Patterns

• It is different to that produced by a single slit or a diffraction grating

The interference pattern produced when white light is diffracted through a double slit

• Each maximum is of roughly equal width
• There are two dark narrow destructive interference fringes on either side
• All other maxima are composed of a   spectrum
• The shortest wavelength (violet / blue) would appear   nearest   to the central maximum
• The longest wavelength (red) would appear   furthest   from the central maximum
• As the maxima move   further away   from the central maximum, the wavelengths of   blue   observed   decrease   and the wavelengths of   red   observed   increase

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Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

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3.2: Young's Double-Slit Interference

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Learning Objectives

By the end of this section, you will be able to:

• Explain the phenomenon of interference
• Define constructive and destructive interference for a double slit

The Dutch physicist Christiaan Huygens (1629–1695) thought that light was a wave, but Isaac Newton did not. Newton thought that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous reputation, his view generally prevailed; the fact that Huygens’s principle worked was not considered direct evidence proving that light is a wave. The acceptance of the wave character of light came many years later in 1801, when the English physicist and physician Thomas Young (1773–1829) demonstrated optical interference with his now-classic double-slit experiment.

If there were not one but two sources of waves, the waves could be made to interfere, as in the case of waves on water (Figure \(\PageIndex{1}\)). If light is an electromagnetic wave, it must therefore exhibit interference effects under appropriate circumstances. In Young’s experiment, sunlight was passed through a pinhole on a board. The emerging beam fell on two pinholes on a second board. The light emanating from the two pinholes then fell on a screen where a pattern of bright and dark spots was observed. This pattern, called fringes, can only be explained through interference, a wave phenomenon.

We can analyze double-slit interference with the help of Figure \(\PageIndex{2}\), which depicts an apparatus analogous to Young’s. Light from a monochromatic source falls on a slit \(S_0\). The light emanating from \(S_0\) is incident on two other slits \(S_1\) and \(S_2\) that are equidistant from \(S_0\). A pattern of interference fringes on the screen is then produced by the light emanating from \(S_1\) and \(S_2\). All slits are assumed to be so narrow that they can be considered secondary point sources for Huygens’ wavelets ( The Nature of Light ). Slits \(S_1\) and \(S_2\) are a distance d apart (\(d≤1\,mm\)), and the distance between the screen and the slits is D(≈1m), which is much greater than d.

Since \(S_0\) is assumed to be a point source of monochromatic light, the secondary Huygens wavelets leaving \(S_1\) and \(S_2\) always maintain a constant phase difference (zero in this case because \(S_1\) and \(S_2\) are equidistant from \(S_0\)) and have the same frequency. The sources \(S_1\) and \(S_2\) are then said to be coherent. By coherent waves, we mean the waves are in phase or have a definite phase relationship. The term incoherent means the waves have random phase relationships, which would be the case if \(S_1\) and \(S_2\) were illuminated by two independent light sources, rather than a single source \(S_0\). Two independent light sources (which may be two separate areas within the same lamp or the Sun) would generally not emit their light in unison, that is, not coherently. Also, because \(S_1\) and \(S_2\) are the same distance from \(S_0\), the amplitudes of the two Huygens wavelets are equal.

Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. In the following discussion, we illustrate the double-slit experiment with monochromatic light (single λ) to clarify the effect. Figure \(\PageIndex{3}\) shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, the slits act as sources of coherent waves and light spreads out as semicircular waves, as shown in Figure \(\PageIndex{1a}\). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure \(\PageIndex{1}\). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double-slit interference pattern, consider how two waves travel from the slits to the screen (Figure \(\PageIndex{5}\)). Each slit is a different distance from a given point on the screen. Thus, different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the path length difference \(\Delta l\) between the two waves is any half-integral number of wavelengths [(1 / 2)λ, (3 / 2)λ, (5 / 2)λ, etc.], then destructive interference occurs. Similarly, if the path length difference is any integral number of wavelengths (λ, 2λ, 3λ, etc.), then constructive interference occurs. These conditions can be expressed as equations:

\[\underbrace{\Delta l = m \lambda}_{\text{constructive interference}} \nonumber \]

for \(m = 0, \, ±1, \, ±2, \, ±3…\)

\[\underbrace{\Delta l = \left(m + \frac{1}{2}\right)\lambda }_{\text{destructuve interference}} \nonumber \]

In Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of monochromatic light used in the experiment. 52.9 589.2 67.90 45.6

The correct option is a 589.2 589.2.

• He placed a screen that had two slits cut into it in front of a monochromatic ( single color ) light.
• The results of Young's Double Slit Experiment should be very different if light is a wave or a particle.
• Let’s look at what the results would be in both situations, and then see how this experiment supports the wave model.

If light is a particle…

We set up our screen and shine a bunch of monochromatic light onto it.

• Imagine it as being almost as though we are spraying paint from a spray can through the openings.
• Since they are little particles they will make a pattern of two exact lines on the viewing screen ( Figure 1 ).

If light is a wave…

If light is a wave, everything starts the same way, but results we get are very different.

• Remember, diffraction is when light passes through a small opening and starts to spread out. This will happen from both openings ( Figure 2 ).
• Where crest meets crest , there will be constructive interference and the waves will make it to the viewing screen as a bright spot .
• Where crest meets trough there will be destructive interference that cancel each other out… a black spot will appear on the screen.
• When this experiment is performed we actually see this, as shown in Figure 3 .

We must conclude that light is made up of waves, since particles can not diffract.

Calculations

When you set up this sort of an apparatus, there is actually a way for you to calculate where the bright lines (called fringes ) will appear.

• There is always a middle line, which is the brightest. We call it the central fringe .
• The central fringe is n = 0 .
• The fringe to either side of the central fringe has an order of n = 1 (the first order fringe ).
• The order of the next fringe out on either side is n = 2 (the second order fringe ).
• And so on, as shown in Figure 4 .

The formula that we will use to figure out problems involving double slit experiments is easy to mix up, so make sure you study it carefully.

λ = wavelength of light used (m) x = distance from central fringe (m) d = distance between the slits (m) n = the order of the fringe L = length from the screen with slits to the viewing screen (m)

It is very easy to mix up the measurements of x, d, and L.

• Make sure to look at Figure 5 and see the different things each is measuring.
• If you mix up x and d it's not so bad, since they are both on top in the formula. If you were to mix them up with L, you would get the wrong answer.
• Almost all questions that you will see for this formula just involve sorting out what each variable is... you might find it helpful to write out a list of givens.

Example 1 : A pair of screens are placed 13.7m apart. A third order fringe is seen on the screen 2.50cm from the central fringe. If the slits were cut 0.0960 cm apart, determine the wavelength of this light. Roughly what colour is it?

Just to make sure you’ve got all the numbers from the question matched with the correct variables… L = 13.7 m n = 3 x = 2.50cm = 0.0250 m d = 0.0960cm = 9.60e-4 m It’s probably a yellow light being used given the wavelength we've measured.

If a white light is used in the double slit experiment, the different colours will be split up on the viewing screen according to their wavelengths.

• The violet end of the spectrum (with the shortest wavelengths) is closer to the central fringe, with the other colours being further away in order.

There is also a version of the formula where you measure the angle between the central fringe and whatever fringe you are measuring.

• The formula works the same way, with the only difference being that we measure the angle instead of x and L.
• Make sure that your calculator is in degree mode before using this version of the formula.

Example 2 : If a yellow light with a wavelength of 540 nm shines on a double slit with the slits cut 0.0100 mm apart, determine what angle you should look away from the central fringe to see the second order fringe?

Do not forget to: Change the wavelength into metres. Change the slit separation into metres. "Second order" is a perfect number and has an infinite number of sig digs.

The Single Slit

A surprising experiment is that you can get the same effect from using a single slit instead of a double slit.

• In Figure 7 , the blue path has to travel further than the red path... if this difference is equal to half a wavelength, they will meet each other out of sync .
• If they meet crest to crest or trough to trough they will be in sync , but if they meet crest to trough they will be out of sync .
• Being in sync will result in constructive interference , while meeting out of sync will result in destructive interference .
• After the first couple of fringes (n = 1 and 2), the edges start getting really fuzzy, so you have a hard time measuring anything.
• The only real difference in calculations is that "d" is now the width of the single opening.
• If two slits work better than one, would more than two slits work better? This is a question that we will answer in the next section.

Example 3 : For a single slit experiment apparatus like the one described above, determine how far from the central fringe the first order violet (λ = 350nm) and red (λ = 700nm) colours will appear if the screen is 10 m away and the slit is 0.050 cm wide.

We need to solve the formula for “x”, the distance from the central fringe. For the violet light… For the red light…

3.1 Young's Double-Slit Interference

Learning objectives.

By the end of this section, you will be able to:

• Explain the phenomenon of interference
• Define constructive and destructive interference for a double slit

The Dutch physicist Christiaan Huygens (1629–1695) thought that light was a wave, but Isaac Newton did not. Newton thought that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous reputation, his view generally prevailed; the fact that Huygens’s principle worked was not considered direct evidence proving that light is a wave. The acceptance of the wave character of light came many years later in 1801, when the English physicist and physician Thomas Young (1773–1829) demonstrated optical interference with his now-classic double-slit experiment.

If there were not one but two sources of waves, the waves could be made to interfere, as in the case of waves on water ( Figure 3.2 ). If light is an electromagnetic wave, it must therefore exhibit interference effects under appropriate circumstances. In Young’s experiment, sunlight was passed through a pinhole on a board. The emerging beam fell on two pinholes on a second board. The light emanating from the two pinholes then fell on a screen where a pattern of bright and dark spots was observed. This pattern, called fringes, can only be explained through interference, a wave phenomenon.

We can analyze double-slit interference with the help of Figure 3.3 , which depicts an apparatus analogous to Young’s. Light from a monochromatic source falls on a slit S 0 S 0 . The light emanating from S 0 S 0 is incident on two other slits S 1 S 1 and S 2 S 2 that are equidistant from S 0 S 0 . A pattern of interference fringes on the screen is then produced by the light emanating from S 1 S 1 and S 2 S 2 . All slits are assumed to be so narrow that they can be considered secondary point sources for Huygens’ wavelets ( The Nature of Light ). Slits S 1 S 1 and S 2 S 2 are a distance d apart ( d ≤ 1 mm d ≤ 1 mm ), and the distance between the screen and the slits is D ( ≈ 1 m ) D ( ≈ 1 m ) , which is much greater than d.

Since S 0 S 0 is assumed to be a point source of monochromatic light, the secondary Huygens wavelets leaving S 1 S 1 and S 2 S 2 always maintain a constant phase difference (zero in this case because S 1 S 1 and S 2 S 2 are equidistant from S 0 S 0 ) and have the same frequency. The sources S 1 S 1 and S 2 S 2 are then said to be coherent. By coherent waves , we mean the waves are in phase or have a definite phase relationship. The term incoherent means the waves have random phase relationships, which would be the case if S 1 S 1 and S 2 S 2 were illuminated by two independent light sources, rather than a single source S 0 S 0 . Two independent light sources (which may be two separate areas within the same lamp or the Sun) would generally not emit their light in unison, that is, not coherently. Also, because S 1 S 1 and S 2 S 2 are the same distance from S 0 S 0 , the amplitudes of the two Huygens wavelets are equal.

Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. In the following discussion, we illustrate the double-slit experiment with monochromatic light (single λ λ ) to clarify the effect. Figure 3.4 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, the slits act as sources of coherent waves and light spreads out as semicircular waves, as shown in Figure 3.5 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3.2 . Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double-slit interference pattern, consider how two waves travel from the slits to the screen ( Figure 3.6 ). Each slit is a different distance from a given point on the screen. Thus, different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the path length difference Δ l Δ l between the two waves is any half-integral number of wavelengths [(1 / 2) λ λ , (3 / 2) λ λ , (5 / 2) λ λ , etc.], then destructive interference occurs. Similarly, if the path length difference is any integral number of wavelengths ( λ λ , 2 λ λ , 3 λ λ , etc.), then constructive interference occurs. These conditions can be expressed as equations:

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• Authors: Samuel J. Ling, Jeff Sanny, William Moebs
• Publisher/website: OpenStax
• Book title: University Physics Volume 3
• Publication date: Sep 29, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-3/pages/3-1-youngs-double-slit-interference

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