In Young's double slit experiment, find fringe width when wavelength and slit distance change

Question

In Young’s double slit experiment, the slit separation is $0.1\,\text{mm}$ and the screen is $1\,\text{m}$ away. If the wavelength of light used is $600\,\text{nm}$ , find the fringe width. What happens to the fringe width if (a) the wavelength is changed to $400\,\text{nm}$ , and (b) the slit separation is doubled?

(CBSE 2024, similar pattern)

Solution — Step by Step

\beta = \frac{\lambda D}{d}

where $\lambda$ = wavelength, $D$ = screen distance, $d$ = slit separation.

\beta = \frac{600 \times 10^{-9} \times 1}{0.1 \times 10^{-3}} = \frac{6 \times 10^{-7}}{10^{-4}} = 6 \times 10^{-3}\,\text{m} = \mathbf{6\,\text{mm}}

\beta' = \frac{400 \times 10^{-9} \times 1}{0.1 \times 10^{-3}} = 4 \times 10^{-3}\,\text{m} = \mathbf{4\,\text{mm}}

The fringe width decreases proportionally with wavelength. Shorter wavelength means narrower fringes.

If $d' = 2d = 0.2\,\text{mm}$ :

\beta'' = \frac{600 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 3 \times 10^{-3}\,\text{m} = \mathbf{3\,\text{mm}}

Doubling the slit separation halves the fringe width.

Why This Works

Fringe width is the distance between two consecutive bright (or dark) fringes. The path difference between waves from the two slits changes by $\lambda$ over one fringe width. Using geometry ( $\sin\theta \approx \tan\theta$ for small angles), this gives $\beta = \lambda D/d$ .

The proportionalities make physical sense: larger wavelength means wider fringes (the waves need more space to shift by one full wavelength); larger slit separation means the two beams converge at steeper angles, compressing the pattern.

Alternative Method — Using path difference

At position $y$ from the central maximum, the path difference is $\Delta = \frac{yd}{D}$ .

For the $m$ th bright fringe: $\frac{y_m d}{D} = m\lambda$ , so $y_m = \frac{m\lambda D}{d}$ .

Fringe width: $\beta = y_{m+1} - y_m = \frac{\lambda D}{d}$ . Same result.

JEE often asks what happens when the setup is immersed in a medium of refractive index $\mu$ . The wavelength in the medium becomes $\lambda/\mu$ , so fringe width becomes $\beta/\mu$ — it decreases. This is a favourite twist question.

Common Mistake

Unit conversion errors are the top scoring killer here. Students forget to convert mm to m or nm to m, leading to answers off by factors of $10^3$ or $10^6$ . Write all quantities in SI units (metres) before substituting. $0.1\,\text{mm} = 10^{-4}\,\text{m}$ and $600\,\text{nm} = 6 \times 10^{-7}\,\text{m}$ . One misplaced power of 10 and the answer is completely wrong.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using path difference

Common Mistake

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