35-2 Young’s Interference Experiment

35-2 Young’s Interference Experiment#

Prompts

  • What is diffraction? Why does light passing through a narrow slit flare out? How does narrowing the slit affect the diffraction?

  • Describe the setup of Young’s double-slit experiment. Why does a single slit come before the two slits? What role does diffraction play?

  • For a point on the screen at angle \(\theta\) from the central axis: what is the path length difference \(\Delta L\) between the two rays? How does \(\Delta L\) relate to constructive vs. destructive interference?

  • Write the conditions for bright fringes and dark fringes in terms of \(d\), \(\theta\), and \(\lambda\). What is the central maximum? The first side maximum? The first minimum?

  • If a transparent plastic strip is placed over one slit, how does it shift the interference pattern? How would you find the thickness needed to shift the \(m=1\) bright fringe to the center?

Lecture Notes#

Overview#

  • Young’s experiment (1801) demonstrated that light is a wave by showing interference—the same phenomenon as overlapping water or sound waves.

  • Light from a single slit illuminates two slits; diffraction causes the light to flare and overlap. The path length difference \(\Delta L\) between the two rays reaching a point on the screen determines whether interference is constructive (bright) or destructive (dark).

  • Bright fringes: \(d\sin\theta = m\lambda\). Dark fringes: \(d\sin\theta = (m + \frac{1}{2})\lambda\). The pattern is symmetric about the central maximum.

Diffraction#

Diffraction is the spreading of a wave when it passes through an opening comparable in size to the wavelength. The narrower the slit, the greater the diffraction—light flares more.

  • Diffraction limits geometrical optics: we cannot form a true “ray” by narrowing slits, because narrowing increases spreading.

  • In Young’s experiment, diffraction at the first slit creates a coherent source; diffraction at the two slits causes the light to overlap and interfere.

Diffraction vs. interference

Diffraction = spreading through an aperture. Interference = combining of overlapping waves. Young’s experiment uses both: diffraction to spread light from the slits, interference to produce the fringe pattern.

Young’s experiment: setup#

Setup: Monochromatic light → slit \(S_0\) (acts as point source) → two slits \(S_1\), \(S_2\) in screen \(B\) → viewing screen \(C\).

  • Light from \(S_0\) diffracts and illuminates both slits. Light from \(S_1\) and \(S_2\) diffracts and overlaps in the region beyond.

  • Where the overlapping waves meet the screen, bright fringes (maxima) and dark fringes (minima) form an interference pattern.

  • The waves leaving the two slits are in phase (they come from the same incident wave). Phase difference at a point on the screen arises only from the path length difference \(\Delta L\).

Path length difference and fringe location#

For a point \(P\) on the screen at angle \(\theta\) from the central axis, the path from \(S_2\) is longer than from \(S_1\) by

(274)#\[ \Delta L = d\sin\theta \]

where \(d\) is the slit separation. This holds when \(D \gg d\) (screen far from slits), so the rays are approximately parallel.

Interference conditions (same logic as section 17-3):

Type

Condition

\(d\sin\theta\)

Bright (maximum)

\(\Delta L = m\lambda\)

\(m\lambda\), \(m = 0, 1, 2, \ldots\)

Dark (minimum)

\(\Delta L = (m + \frac{1}{2})\lambda\)

\((m + \frac{1}{2})\lambda\), \(m = 0, 1, 2, \ldots\)
(275)#\[ d\sin\theta = m\lambda \quad \text{(bright)} \]
(276)#\[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \quad \text{(dark)} \]

  • Central maximum (\(m = 0\)): \(\theta = 0\), \(\Delta L = 0\)—waves arrive in phase.

  • First side maxima (\(m = 1\)): \(d\sin\theta = \lambda\)—one wavelength path difference.

  • First minima (\(m = 0\) in dark formula): \(d\sin\theta = \lambda/2\)—half-wavelength path difference.

Distance on the screen#

When \(\theta\) is small, \(\tan\theta \approx \sin\theta \approx \theta\) (in radians). The vertical distance \(y\) from the center of the pattern to a point at angle \(\theta\) is

(277)#\[ y = D\tan\theta \approx D\theta \]

where \(D\) is the slit-to-screen distance. For the \(m\)th bright fringe:

(278)#\[ y_m = \frac{m\lambda D}{d} \]

Fringe spacing (distance between adjacent bright fringes):

(279)#\[ \Delta y = \frac{\lambda D}{d} \]

  • Increasing \(\lambda\) or \(D\) → fringes spread farther apart.

  • Increasing \(d\) → fringes squeeze together.

Important

The approximation \(\sin\theta \approx \theta\) holds for small angles. For large \(\theta\), use the exact relations \(d\sin\theta = m\lambda\) and \(y = D\tan\theta\).

Poll: Fringe spacing—effect of slit separation

A double-slit interference pattern is produced with monochromatic light (600 nm). What happens to the fringe separation \(\Delta y\) if you decrease the slit separation \(d\)?

  • (A) \(\Delta y\) increases

  • (B) \(\Delta y\) decreases

  • (C) \(\Delta y\) stays the same

Poll: Fringe spacing—effect of index (e.g., tank of liquid)

Same double-slit setup. You fill the space between slits and screen with a liquid (\(n > 1\)). What happens to \(\Delta y\)?

  • (A) \(\Delta y\) increases

  • (B) \(\Delta y\) decreases

  • (C) \(\Delta y\) stays the same

Poll: Fringe spacing—effect of screen distance

Same setup. What happens to \(\Delta y\) if you decrease the slit-to-screen distance \(D\)?

  • (A) \(\Delta y\) increases

  • (B) \(\Delta y\) decreases

  • (C) \(\Delta y\) stays the same

Poll: Fringe spacing—effect of wavelength

Same setup with 600 nm light. You switch to 632 nm light. What happens to \(\Delta y\)?

  • (A) \(\Delta y\) increases

  • (B) \(\Delta y\) decreases

  • (C) \(\Delta y\) stays the same

Example: Find slit separation from measured fringe spacing

A double-slit pattern with 600 nm light is produced on a screen 3.0 m away. The distance between the \(m\)th and \((m+8)\)th bright fringes is 24 cm. Find the slit separation \(d\).

Solution: \(\Delta y = 24\,\text{cm}/8 = 3.0\) cm. From \(\Delta y = \lambda D/d\), we get \(d = \lambda D/\Delta y = (600\times10^{-9})(3.0)/(0.030) = 60\,\mu\text{m}\).

Plastic over one slit#

If a transparent material (index \(n\), thickness \(L\)) covers one slit, light through that slit accumulates extra phase because \(\lambda_n = \lambda/n\)—more wavelengths fit in the same length.

To shift the \(m\)th bright fringe to the center: the phase difference from the plastic must compensate for the path length difference that originally placed that fringe off-center. For the \(m=1\) fringe (one wavelength path difference) moved to center:

(280)#\[ (n - 1)\frac{L}{\lambda} = 1 \quad \Rightarrow \quad L = \frac{\lambda}{n-1} \]

  • Place the plastic over the slit whose ray had the longer path to the fringe you want at center. The plastic adds phase to that ray (more wavelengths fit in length \(L\) because \(\lambda_n = \lambda/n\)).

  • For the lower \(m=1\) fringe: ray \(r_2\) (top slit) had \(\Delta L = \lambda\) more than \(r_1\). To put that fringe at center (\(\Delta L = 0\)), we need the ray through the top slit to gain one extra wavelength of phase—place plastic over the top slit.

Example: Fringe spacing

Light of wavelength 546 nm, slit separation \(d = 0.12\) mm, screen distance \(D = 55\) cm. Fringe spacing \(\Delta y = \lambda D/d = (546\times10^{-9})(0.55)/(0.12\times10^{-3}) \approx 2.5\) mm.

Example: Plastic over one slit

To shift the \(m=1\) bright fringe to center using plastic with \(n = 1.50\) and \(\lambda = 600\) nm: \(L = \lambda/(n-1) = 600\,\text{nm}/0.50 = 1.2\,\mu\text{m}\). Place plastic over the slit whose ray had the longer path to the original \(m=1\) fringe.

Summary#

  • Diffraction: waves spread through narrow openings; narrower slit → more spreading.

  • Young’s experiment: single slit → two slits → interference pattern from path length difference.

  • Bright: \(d\sin\theta = m\lambda\). Dark: \(d\sin\theta = (m + \frac{1}{2})\lambda\).

  • Screen position: \(y \approx D\theta\); fringe spacing \(\Delta y = \lambda D/d\).

  • Plastic over one slit: adds phase via \(\lambda_n = \lambda/n\); use \((n-1)L/\lambda\) to find thickness for desired shift.