ARCHETYPE: YDSE & Fringe-Width Numericals

Young's double slit is not one exam question — it is a factory that stamps out dozens of them from a single die. Give a paper-setter two slits, a screen and a wavelength, and they can ask for the fringe spacing, the position of the fourth bright band, the shift when a thin glass flake is dropped over one slit, the squeeze when the whole apparatus is dunked in water, or the wavelength that produced a measured pattern. Every one of these is the same physics — path difference across a small angle — dressed in different clothes. Learn the machine once and the entire family becomes fill-in-the-blanks. This is a numerical gym, so the setup is the article. At its centre sits one formula you already met: the fringe spacing $\beta = \lambda D / d$ , the uniform centre-to-centre distance between neighbouring bright (or dark) bands, where $d$ is the slit separation and $D$ the slit-to-screen distance. Everything else is a variation on it. Bright fringes sit at $y_n = n\lambda D/d$ , dark ones halfway between. Slip a transparent slab of index $\mu$ and thickness $t$ into one beam and the whole pattern slides by $N = (\mu-1)t/\lambda$ fringes toward that side. Immerse the apparatus in a liquid of index $n$ and every fringe shrinks to $\beta' = \beta/n$ , because the wavelength itself shrinks. The trap that catches half the class is a single letter: $d$ is the slit separation, $a$ is the slit width — one governs interference (this article), the other diffraction (the next gym). Keep them straight, carry four significant figures, and the YDSE questions on any Indian board paper, on AP Physics 2, and on A-Level 9702 turn from feared to routine. This article drills the master recipe and the four variants examiners actually pose, with every number computed to the last digit.

Key takeaways

The engine of every YDSE numerical is the fringe spacing $\beta = \lambda D/d$ — uniform spacing of bright and dark bands, with $d$ the slit separation and $D$ the slit-to-screen distance.
Bright fringes at $y_n = n\lambda D/d$ ; dark fringes at $y_n = (n-\tfrac12)\lambda D/d$ — halfway between the bright ones.
Slab shift: a plate of index $\mu$ , thickness $t$ in one path moves the pattern by $N = (\mu-1)t/\lambda$ fringes toward the slab (extra optical path $(\mu-1)t$ ).
Immersion: dunking the whole apparatus in a liquid of index $n$ shrinks every fringe to $\beta' = \beta/n$ .
Find $\lambda$ from a measured fringe set: $\lambda = \beta d/D$ . And beware missing orders where an interference maximum lands on a diffraction minimum ( $d/a$ a whole number).
Key idea: one relation, $\beta = \lambda D/d$ , plus the slab shift $N = (\mu-1)t/\lambda$ and immersion $\beta' = \beta/n$ , solve the entire YDSE numerical family — provided you never confuse the separation $d$ with the width $a$ .

The core insight

Every double-slit numerical reduces to the small-angle path-difference geometry $\Delta = dy/D$ with bright fringes at $\Delta = n\lambda$ , giving the uniform fringe spacing $\beta = \lambda D/d$ ; a slab shifts the pattern by $N = (\mu-1)t/\lambda$ fringes and immersion divides $\beta$ by the liquid's index, so the whole family is one formula and its variations.

Where this lives today

The double slit grew up into the interferometer, and interferometers now measure the world's finest lengths. A Fabry–Pérot cavity — the sharpened double slit — sits inside every stabilized laser and every optical atomic clock, and its cousins gauge the flatness of mirrors to a fraction of a wavelength. The same fringe-counting logic reads out the mirror positions in LIGO, the gravitational-wave detector that measured spacetime ripples a thousandth the width of a proton by watching interference fringes shift. Wavelength-division multiplexing packs dozens of laser colours down one optical fibre using interference filters descended from these ideas, and the "slab shift" you just computed is the working principle of every phase-contrast microscope and optical thickness gauge, which read a transparent sample's thickness from exactly the $(\mu-1)t$ path it adds. Young's two slits, drilled as exam arithmetic here, are the ancestor of a whole industry of precision measurement.

The setup

Geometry first, because in a numerical gym the geometry is the physics. A single monochromatic source lights two narrow slits a distance $d$ apart. Each slit re-radiates as a coherent source (they share one parent beam), and on a screen a distance $D$ away — with $D \gg d$ — the two beams overlap and interfere. To reach a point $y$ above the centre line, the light from the far slit travels a little farther than the light from the near slit; for the small angles of a real bench that extra path is

\Delta = \frac{d\,y}{D}. \qquad (1)

where: $\Delta$ — the path difference between the two beams at the screen point $y$ ; $d$ — slit separation; $D$ — slit-to-screen distance; $y$ — height above the central axis. A bright band forms wherever the two beams arrive in step, $\Delta = n\lambda$ ; a dark band wherever they arrive exactly out of step, $\Delta = (n-\tfrac12)\lambda$ . That is the whole engine — turn the crank and every fringe position falls out.

Figure

The double-slit bench with fringe spacing beta equal to lambda D over d, and a thin glass slab over one slit sliding the whole pattern toward that side

The master relations

From $\Delta = dy/D$ and the bright condition $\Delta = n\lambda$ , the $n$ -th bright fringe sits at $y_n = n\lambda D/d$ , and consecutive fringes are one $\lambda D/d$ apart — the fringe spacing:

\beta = \frac{\lambda D}{d}, \qquad y_n^{\text{bright}} = n\frac{\lambda D}{d}, \qquad y_n^{\text{dark}} = \left(n-\tfrac12\right)\frac{\lambda D}{d}. \qquad (2)

where: $\beta$ — the uniform centre-to-centre fringe spacing; $n = 0, 1, 2, \dots$ orders the bright fringes ( $n=0$ is the central bright band). A quick reference set, which we will reuse: $\lambda = 600\ \mathrm{nm}$ , $d = 1\ \mathrm{mm}$ , $D = 1\ \mathrm{m}$ give $\beta = 0.6\ \mathrm{mm}$ , so bright fringes fall at $y_1 = 0.6\ \mathrm{mm}$ , $y_2 = 1.2\ \mathrm{mm}$ , $y_3 = 1.8\ \mathrm{mm}$ , and dark fringes at $y_1 = 0.3\ \mathrm{mm}$ , $y_2 = 0.9\ \mathrm{mm}$ . The angular fringe spacing is $\theta = \lambda/d = 6\times10^{-4}\ \mathrm{rad}$ . Two more relations complete the toolkit. A transparent slab of index $\mu$ and thickness $t$ placed over one slit adds an extra optical path $(\mu-1)t$ to that beam (light of geometric length $t$ now takes an optical length $\mu t$ , a surplus of $(\mu-1)t$ over the air it replaced). Dividing by $\lambda$ gives the shift in whole fringes:

N = \frac{(\mu-1)t}{\lambda}, \qquad \beta' = \frac{\beta}{n_{\text{liq}}}. \qquad (3)

where: $N$ — the number of fringes the pattern slides toward the slab side; $\beta'$ — the fringe spacing after immersing the whole apparatus in a liquid of index $n_{\text{liq}}$ , which shrinks the wavelength (and hence every fringe) by that factor.

Derivation AThe fringe spacing β = λD/d from the small-angle path difference

The goal is to obtain the master formula $\beta = \lambda D/d$ and the fringe positions of equation (2) from nothing but the path difference $\Delta = dy/D$ and the bright-fringe condition $\Delta = n\lambda$ . The strategy is pure small-angle geometry — no calculus, no Fourier transform of the slits. Watch the roles of $d$ (separation) and $y$ (screen height): they combine into the path difference; the wavelength then sets which heights are bright.

Step 1. Write the path difference to a screen point $y$ . With the screen far away ( $D \gg d$ ) the two rays are nearly parallel, and the far slit's extra path is $\Delta = d\sin\theta \approx d\tan\theta = dy/D$ for the small angle $\theta$ . This is equation (1).

Step 2. Impose the bright condition. Constructive interference needs $\Delta = n\lambda$ , so $dy_n/D = n\lambda$ , giving $y_n = n\lambda D/d$ — the $n$ -th bright fringe. The dark condition $\Delta = (n-\tfrac12)\lambda$ gives the dark fringes at $y_n = (n-\tfrac12)\lambda D/d$ , exactly halfway between.

Step 3. Subtract consecutive bright fringes for the spacing: $\beta = y_{n+1} - y_n = (n+1)\lambda D/d - n\lambda D/d = \lambda D/d$ . The spacing is uniform (independent of $n$ ) and identical for bright and dark bands.

Step 4. Read the reference numbers. $\lambda = 600\ \mathrm{nm}$ , $d = 1\ \mathrm{mm}$ , $D = 1\ \mathrm{m}$ : $\beta = (600\times10^{-9})(1)/(10^{-3}) = 6\times10^{-4}\ \mathrm{m} = 0.6\ \mathrm{mm}$ . Bright fringes at $0.6, 1.2, 1.8\ \mathrm{mm}$ ; dark at $0.3, 0.9\ \mathrm{mm}$ . Verified: $\beta = 0.6\ \mathrm{mm}$ and the bright/dark fringe positions are recomputed independently in verify_hs2-ydse-problems.py.

Go deeper · Derivation B · 1 minThe slab shift N = (μ−1)t/λ from the extra optical path

Derivation BThe slab shift N = (μ−1)t/λ from the extra optical path

The goal is the plate-shift formula $N = (\mu-1)t/\lambda$ : how many fringes the whole pattern slides when a thin transparent slab is dropped over one slit. The strategy is to compute the extra optical path the slab introduces, then convert that path into a fringe count. The key idea, carried into the next capstone, is that optical path equals $\mu \times$ geometric path, and one wavelength of extra optical path shifts the pattern by exactly one fringe.

Step 1. Find the extra optical path. Where the beam used to cross thickness $t$ of air (optical length $t$ ), it now crosses thickness $t$ of slab (optical length $\mu t$ ). The surplus over what was there before is $(\mu t) - (t) = (\mu-1)t$ .

Step 2. Convert to fringes. The pattern shifts by one whole fringe for every extra wavelength of optical path (a $2\pi$ of phase). So the shift in fringes is the extra path divided by the wavelength: $N = (\mu-1)t/\lambda$ .

Step 3. Note the direction. The slab slows the light on its side, so that beam falls behind; the pattern shifts toward the slab side to restore the zero-path-difference point on that side. Crucially, the shift $N$ does not depend on $d$ or $D$ — it is set entirely by the slab.

Step 4. Plug numbers. $\mu = 1.5$ , $t = 2\ \mu\mathrm{m}$ , $\lambda = 600\ \mathrm{nm}$ : $N = (1.5-1)(2\times10^{-6})/(600\times10^{-9}) = (0.5)(2\times10^{-6})/(6\times10^{-7}) = 1.667$ fringes. Verified: extra optical path $(\mu-1)t$ and the shift $N = 1.667$ fringes are recomputed in verify_hs2-ydse-problems.py.

Figure

A straight-line graph of fringe spacing beta against one over the slit separation, passing through the origin with slope lambda times D

The variants examiners actually pose

Find the wavelength. Rearrange equation (2): $\lambda = \beta d/D$ . Measure the fringe spacing (often by counting several fringes across a known distance) and back out $\lambda$ — this is how Young first measured the wavelength of light.

The slab shift. Use $N = (\mu-1)t/\lambda$ . The pattern moves toward the slab; the shift is independent of $d$ and $D$ . A shift of $N$ fringes moves the pattern a distance $N\beta$ on the screen.

Immersion. Put the whole apparatus in a liquid: the wavelength becomes $\lambda/n_{\text{liq}}$ , so $\beta' = \beta/n_{\text{liq}}$ . Every fringe crowds inward by the index of the liquid.

Missing orders. Real slits have a width $a$ as well as a separation $d$ . Wherever an interference maximum ( $d\sin\theta = n\lambda$ ) coincides with a diffraction minimum ( $a\sin\theta = m\lambda$ ), the bright fringe is snuffed out — a missing order. This happens when $d/a = n/m$ is a whole-number ratio; if $d/a = 3$ , then orders $n = 3, 6, 9,\dots$ vanish.

Trap table and limiting cases

pitfall or case	what actually happens	the fix
$d$ (separation) vs $a$ (width)	$d$ sets fringe spacing $\beta = \lambda D/d$ ; $a$ sets diffraction	label them before touching numbers
slab shift uses $\mu t$ ?	no — the surplus is $(\mu-1)t$ , so $N = (\mu-1)t/\lambda$	subtract the air the slab replaced
immersion effect	$\beta' = \beta/n_{\text{liq}}$ — fringes shrink	wavelength shrinks in the liquid
dark fringe position	$(n-\tfrac12)\lambda D/d$ , halfway between bright	not $n\lambda D/d$
$d \to$ large	$\beta \to 0$ , fringes merge	wide separation gives fine, unresolvable fringes
$D \to$ large	$\beta \to$ large	a distant screen magnifies the pattern
missing order	maximum on a diffraction minimum ( $d/a$ integer)	count which orders vanish

Common trapThe separation d and the width a are different letters for a reason

The single costliest habit in this whole gym is sliding between the slit separation $d$ and the slit width $a$ . Fringe spacing is $\beta = \lambda D/d$ — the separation $d$ , the distance between the two slits' centres. The slit width $a$ belongs to diffraction (the next article): it sets how broad the single-slit envelope is and where diffraction minima fall. They enter different formulas and often differ by a factor of ten in the same question. A second, related slip: the slab shift $N = (\mu-1)t/\lambda$ uses $(\mu-1)t$ , the extra optical path over the air the slab displaced, never the full $\mu t$ . Write down which symbol is which the moment you read the question, and check that your $\beta$ used a separation and your shift used a $(\mu-1)$ .

Worked examples

1 · Board level — fringe spacing from the data (CBSE and NEET pattern)

A double-slit apparatus uses sodium light of wavelength $589\ \mathrm{nm}$ , slit separation $d = 0.2\ \mathrm{mm}$ , and a screen $D = 1.0\ \mathrm{m}$ away. Find the fringe spacing.

Step 1. Identify the letters: $\lambda = 589\ \mathrm{nm}$ , $d = 0.2\ \mathrm{mm} = 2\times10^{-4}\ \mathrm{m}$ (separation, not width), $D = 1.0\ \mathrm{m}$ .

Step 2. Apply equation (2): $\beta = \lambda D/d = (589\times10^{-9})(1.0)/(2\times10^{-4})$ .

Step 3. Compute: $\beta = (589\times10^{-9})/(2\times10^{-4}) = 2.945\times10^{-3}\ \mathrm{m} = \boxed{2.945\ \mathrm{mm}}$ .

Step 4. Sense-check: about $3\ \mathrm{mm}$ between bright bands is comfortably visible, as it should be for a benchtop demonstration with a fine slit pair.

2 · Board level — find the wavelength (ICSE and JEE pattern)

In a double-slit experiment with $d = 0.3\ \mathrm{mm}$ and $D = 1.0\ \mathrm{m}$ , the tenth bright fringe from the centre lies $2.0\ \mathrm{cm}$ from it. Find the wavelength of the light.

Step 1. The tenth bright fringe is at $y_{10} = 10\beta$ , so $\beta = y_{10}/10 = 2.0\ \mathrm{cm}/10 = 0.20\ \mathrm{cm} = 2.0\times10^{-3}\ \mathrm{m}$ .

Step 2. Invert equation (2) for the wavelength: $\lambda = \beta d/D$ .

Step 3. Compute: $\lambda = (2.0\times10^{-3})(0.3\times10^{-3})/(1.0) = 6.0\times10^{-7}\ \mathrm{m} = \boxed{600\ \mathrm{nm}}$ .

Step 4. A wavelength of $600\ \mathrm{nm}$ is orange-red — a plausible source. This is exactly Young's method: measure a fringe spacing, back out $\lambda$ .

3 · Competitive level — the slab shift (JEE-Advanced and NEET pattern)

A thin glass slab ( $\mu = 1.5$ ) of thickness $t = 2\ \mu\mathrm{m}$ is placed over one slit of a double-slit apparatus lit by $600\ \mathrm{nm}$ light. By how many fringes does the pattern shift, and in which direction?

Step 1. The slab adds extra optical path $(\mu-1)t = (0.5)(2\times10^{-6}) = 1.0\times10^{-6}\ \mathrm{m} = 1\ \mu\mathrm{m}$ to that beam.

Step 2. Convert to fringes with equation (3): $N = (\mu-1)t/\lambda = (1.0\times10^{-6})/(600\times10^{-9})$ .

Step 3. Compute: $N = (1.0\times10^{-6})/(6.0\times10^{-7}) = \boxed{1.667\ \text{fringes}}$ , and the pattern shifts toward the slab side (that beam is delayed).

Step 4. Note what does not appear: neither $d$ nor $D$ entered — the shift depends only on the slab. On the screen the pattern moves a distance $N\beta = 1.667\beta$ , whatever $\beta$ happens to be. Timing note: about 60 seconds once you see $N = (\mu-1)t/\lambda$ . The only trap is using $\mu t$ instead of $(\mu-1)t$ — that would give $5.0$ fringes, more than triple the true answer.

4 · Competitive level — immersion in water (JEE-Main and NEET pattern)

A double-slit pattern has fringe spacing $\beta = 0.6\ \mathrm{mm}$ in air. The entire apparatus is now immersed in water ( $n = 1.33$ ). Find the new fringe spacing, and state what happens to the number of fringes visible in a fixed region of the screen.

Step 1. Immersion shrinks the wavelength to $\lambda/n$ , so the spacing shrinks by the same factor: $\beta' = \beta/n = 0.6\ \mathrm{mm}/1.33$ .

Step 2. Compute: $\beta' = (0.6\times10^{-3})/1.33 = 4.51\times10^{-4}\ \mathrm{m} = \boxed{0.451\ \mathrm{mm}}$ .

Step 3. The fringes crowd closer by the factor $1.33$ , so about $33\%$ more fringes fit into any fixed width of screen.

Step 4. Sense-check the direction: water slows light and shrinks its wavelength, and a smaller $\lambda$ gives a smaller $\beta = \lambda D/d$ — fringes must move inward, which they do. Only the wavelength changed; $d$ and $D$ are the same bench. Timing note: about 45 seconds — immersion is a single division by the liquid's index. Do not also change $d$ or $D$ ; the apparatus is unchanged, only the medium.

Exam traps

Confusing $d$ (separation) with $a$ (width). Fringe spacing uses the separation, $\beta = \lambda D/d$ . The width $a$ governs diffraction and appears in the next gym. Mixing them is the single biggest mark-loser here; write both symbols down explicitly.

Using $\mu t$ for the slab shift. The shift is $N = (\mu-1)t/\lambda$ , using the extra path over the displaced air. Using $\mu t$ inflates the answer (here from $1.667$ to $5.0$ fringes). Always subtract the air the slab replaced.

Placing dark fringes wrong. Dark fringes sit at $(n-\tfrac12)\lambda D/d$ , exactly halfway between bright ones — not at $n\lambda D/d$ . The central point ( $n=0$ ) is bright.

Changing the wrong quantity on immersion. Immersion changes the wavelength only, so divide $\beta$ by the liquid's index; leave $d$ and $D$ alone. Students sometimes wrongly scale $D$ or forget the effect entirely.

Unit slips. Keep everything in SI before dividing: $\mathrm{mm}$ and $\mu\mathrm{m}$ and $\mathrm{nm}$ in the same expression cause order-of-magnitude errors. Convert first, then compute, and carry four significant figures.

Board scope. Double-slit numericals are firmly examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET, and also on AP Physics 2 (which does $\lambda = ax/D$ -style problems) and A-Level 9702. The slab-shift and immersion numericals are India-facing specialities — AP and A-Level do the core $\beta = \lambda D/d$ work but rarely the plate-shift bookkeeping, so weight your practice to your board.

Origins · 1 minCharles Fabry and how this was discovered

Charles Fabry

1867–1945 · Marseille and Paris

Charles Fabry was the French physicist who turned interference fringes into a precision measuring instrument. With his colleague Alfred Pérot he built, in the late 1890s, the Fabry–Pérot interferometer — two parallel partially-silvered mirrors that bounce light back and forth so that thousands of beams interfere, producing fringes far sharper than Young's two-beam bands. It is the high-resolution descendant of the double slit, and the fringe-order-versus-path-difference bookkeeping that Fabry mastered is exactly the skill this article drills. Fabry used his interferometer to measure wavelengths to unprecedented accuracy and, in a celebrated 1913 study, helped establish that ozone in the upper atmosphere absorbs the Sun's ultraviolet — a discovery of lasting consequence. He became the first head of the Institut d'Optique in Paris and a member of the Académie des sciences. Credit for the interferometer is properly shared with Pérot, his equal partner in its design.

History

1801

— Thomas Young demonstrates two-slit interference of light and uses it to measure the wavelength.

1819

— Augustin-Jean Fresnel and François Arago establish the laws of interference of polarized light, cementing the wave theory.

1868

— Anders Ångström tabulates solar spectral wavelengths to a ten-billionth of a metre, giving fringe metrology its scale.

1897

— Charles Fabry and Alfred Pérot build their multiple-beam interferometer, the precision heir to Young's two beams.

1960

— the laser provides a bright, highly coherent source that turns the double-slit experiment into a routine lecture demonstration.

Going further

Texts. NCERT Physics Class 12, Ch. 10 ("Wave Optics"), the double-slit sections and exercises · H. C. Verma, Concepts of Physics Vol. 1, "Light Waves," worked interference problems · Irodov, Problems in General Physics, interference set (for the slab-shift and immersion variants) · Cambridge A-Level 9702 (double-slit, $\lambda = ax/D$ ).

Provenance. Standard exam-archive problems. Recompute: reference $\beta = 0.6\ \mathrm{mm}$ ( $600\ \mathrm{nm}$ , $1\ \mathrm{mm}$ , $1\ \mathrm{m}$ ), bright $0.6/1.2/1.8\ \mathrm{mm}$ , dark $0.3/0.9\ \mathrm{mm}$ , angular $6\times10^{-4}\ \mathrm{rad}$ ; Board 1 $\beta = 2.945\ \mathrm{mm}$ ( $589\ \mathrm{nm}$ , $0.2\ \mathrm{mm}$ , $1\ \mathrm{m}$ ); Board 2 $\beta = 2.0\ \mathrm{mm} \Rightarrow \lambda = 600\ \mathrm{nm}$ ( $0.3\ \mathrm{mm}$ , $1\ \mathrm{m}$ ); slab shift $N = (0.5)(2\ \mu\mathrm{m})/(600\ \mathrm{nm}) = 1.667$ ; immersion $\beta' = 0.6\ \mathrm{mm}/1.33 = 0.451\ \mathrm{mm}$ ; missing orders at $d/a$ integer (e.g. $d/a = 3$ ); $\beta$ -vs- $1/d$ slope $\lambda D = 6\times10^{-7}\ \mathrm{m}^2$ .

Prerequisites. Young's Double-Slit Experiment — the concept article that derives $\beta = \lambda D/d$ this gym uses. Leads to. the university seams Young's Double-Slit Experiment and Optical Path Difference and Phase Retardation.

Go deeper (UG). The full treatment of coherence length, source-slit width and fringe visibility — why real fringes fade at high order — is developed in Young's Double-Slit Experiment, and the optical-path bookkeeping behind the slab shift is generalised in Optical Path Difference and Phase Retardation. Enrichment only.

How do I keep the slit separation d and the slit width a straight?

Draw the apparatus and label both before you touch a number. The separation $d$ is the distance between the centres of the two slits — it sets the interference fringe spacing $\beta = \lambda D/d$ , the topic of this gym. The width $a$ is how wide each individual slit is — it sets the single-slit diffraction envelope, the topic of the next gym. In a typical question $d$ might be $0.2\ \mathrm{mm}$ while $a$ is $0.02\ \mathrm{mm}$ , ten times smaller. If a problem gives you both, the separation goes into $\beta$ and the width goes into any diffraction-minimum or missing-order calculation. Mislabel them and every number downstream is wrong, which is why examiners give both.

Why does a glass slab shift the pattern rather than just dimming it?

Because it delays one beam. The slab replaces a thickness $t$ of air with a thickness $t$ of glass, and light travels more slowly in glass, so that beam arrives with an extra optical path $(\mu-1)t$ — as if its slit had moved. To restore the point of zero path difference, the whole fringe pattern slides toward the slab side by $N = (\mu-1)t/\lambda$ fringes. The slab does dim the beam slightly too, but the headline effect examiners test is the shift. Note that the shift is set entirely by the slab's $\mu$ and $t$ and the wavelength — the geometry $d$ and $D$ do not enter, which is a useful check on your working. This plate-shift bookkeeping (and the immersion variant) is an India-facing speciality (CBSE, ICSE, JEE, NEET); AP Physics 2 and A-Level 9702 examine the core $\beta = \lambda D/d$ family but rarely the slab shift.

When the apparatus is immersed in water, why do the fringes get closer, not farther apart?

Because the wavelength shrinks. The fringe spacing is $\beta = \lambda D/d$ , directly proportional to the wavelength. Inside water the wavelength becomes $\lambda/n$ (the frequency is fixed by the source, but light slows, so the wavelength contracts), and therefore $\beta' = \beta/n$ . With $n = 1.33$ the fringes crowd $33\%$ closer. Nothing mechanical changed — the same slits, the same screen distance — only the medium the light travels through, and that alone controls the wavelength. This is a favourite one-step problem precisely because students expect immersion to do something complicated, when it is a single division by the liquid's index.

What are missing orders, and how do I find them?

A missing order is a bright interference fringe that never appears because it lands exactly on a diffraction minimum from the finite slit width. Interference maxima occur at $d\sin\theta = n\lambda$ ; single-slit diffraction minima occur at $a\sin\theta = m\lambda$ . Where these coincide, the interference maximum has zero light to brighten, so it vanishes. Dividing the two conditions, the coincidence happens when $n/m = d/a$ . So if $d/a = 3$ , the orders $n = 3, 6, 9,\dots$ are missing. To find them, form the ratio $d/a$ ; if it is a whole number, that number and its multiples are your missing orders. This ties the two gyms together — the separation sets the fringes, the width erases some of them.

How many significant figures should I carry, and where do people lose marks?

Carry four through the working and present two or three. The classic loss is rounding $\beta$ or an intermediate optical path early and then multiplying, which shifts the final digit and misses a "to three significant figures" answer box. Store the unrounded value in your calculator. The second loss is units: convert millimetres, micrometres and nanometres to metres before dividing, because a stray factor of $10^3$ is an instant zero. The third is copying the wrong constant — sodium is $589\ \mathrm{nm}$ , the common "red-orange" worked value is $600\ \mathrm{nm}$ , green is $550\ \mathrm{nm}$ ; use the wavelength the question actually states.

Is this the same as the concept article on Young's double slit?

It shares the physics but not the purpose. The concept article derives why fringes form — coherence, path difference, the meaning of $\beta = \lambda D/d$ — and proves the formulas. This article assumes those results and drills using them fast and correctly: identifying $d$ versus $a$ , applying the slab shift and immersion, finding $\lambda$ from data, and dodging the traps under exam time. Think of the concept article as the theory lecture and this as the problem class. If a formula here feels unmotivated, revisit the concept article; if you understand the physics but keep making arithmetic and setup slips, this gym is where you fix them.