Question
In Young's double slit experiment, monochromatic light of wavelength 600 nm is used. The slits are separated by 0.5 mm and the screen is placed 1.5 m away. Find the fringe width. If the slit separation is halved and the screen distance is doubled, what happens to the fringe width?
Solution — Step by Step
Write the fringe width formula
The fringe width (β) is the distance between two consecutive bright or dark fringes. The formula is:
Here λ is wavelength, D is the distance from slits to screen, and d is the slit separation. All three live in this one formula — learn it cold.
Convert units before plugging in
Always convert to SI units first. This is where marks leak.
- λ = 600 nm = 600 × 10⁻⁹ m
- d = 0.5 mm = 0.5 × 10⁻³ m = 5 × 10⁻⁴ m
- D = 1.5 m
Calculate fringe width
β = 1.8 mm
Handle the changed conditions
New conditions: d' = d/2, D' = 2D. Substitute into the formula:
So β' = 4 × 1.8 mm = 7.2 mm. The fringe width becomes 4 times the original.
State the proportionality clearly
This is the real takeaway. Since , halving d doubles β, and doubling D doubles β again. Combined effect: 4× increase. JEE loves asking this as a ratio problem.
Why This Works
The fringe pattern forms because light from two coherent sources (the slits) takes different path lengths to reach any point on the screen. At points where the path difference is an integer multiple of λ, waves arrive in phase — bright fringe. Half-integer multiples give dark fringes.
The gap between consecutive bright fringes is β = λD/d because we need the path difference to change by exactly λ as we move from one fringe to the next. Moving the screen farther (larger D) stretches the geometry, so fringes spread apart. Bringing slits closer (smaller d) means a smaller angle is needed to accumulate path difference λ, but the fringes are farther apart on screen — counterintuitive at first glance.
Think of d and D as opposing controls: d is in the denominator (bigger d → tighter fringes), D is in the numerator (bigger D → wider fringes). This relationship appears in NEET and JEE Main almost every year.
Alternative Method — Using Path Difference Geometry
For the nth bright fringe, the condition is: path difference = nλ.
From geometry, the position of nth bright fringe on screen is:
Fringe width is the gap between nth and (n+1)th fringe:
Same result, derived from first principles. This approach is useful when the examiner asks you to derive the formula rather than just apply it — CBSE 12 board frequently asks 3-mark derivation questions on this exact point.
💡 Expert Tip
For quick ratio problems in JEE Main, treat β as a simple product: β ∝ λ × D × (1/d). Any change to these three quantities multiplies β by that factor. If λ doubles, D triples, and d stays same — new β is 6 times original. No formula re-derivation needed.
Common Mistake
⚠️ Common Mistake
Forgetting to convert mm to m before calculating. Students write d = 0.5 and D = 1.5, then get β = 1800 m — a nonsense answer they don't catch under exam pressure. The fix: always write units next to every value you plug in. If units don't cancel to metres, something is wrong. This exact error costs 4-mark questions in JEE Main every session.
A second trap: thinking fringe width changes when we shift the entire setup (move both slits and screen together keeping D constant). It doesn't — β depends only on λ, D, and d, not on the absolute position of anything.