White light is passed through a double slit and interference is observed on a screen 1.5 m away. The separation between the slits is 0.3 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringes. the difference in wavelengths of red and violet light is ................ nm.

Wave Optics PYQ — JEE Main 2021 · 26 Aug · Shift 1 · Q69 with Solution

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Correct answer: 300

For Young’s double slit experiment, the position of the $m$ -th bright fringe is

$y_m = m\frac{\lambda D}{d}$

Here, for the first fringe, $m=1$ , so

$y = \frac{\lambda D}{d}$

Thus,

$\lambda = \frac{yd}{D}$

We need the difference in wavelengths of red and violet light:

$\Delta \lambda = \lambda_r - \lambda_v = \frac{(y_r-y_v)d}{D}$

Screen distance: $D = 1.5\,\text{m}$
Slit separation: $d = 0.3\,\text{mm} = 3.0\times10^{-4}\,\text{m}$
First violet fringe position: $y_v = 2.0\,\text{mm} = 2.0\times10^{-3}\,\text{m}$
First red fringe position: $y_r = 3.5\,\text{mm} = 3.5\times10^{-3}\,\text{m}$

So,

$y_r-y_v = (3.5-2.0)\times10^{-3} = 1.5\times10^{-3}\,\text{m}$

$\Delta \lambda = \frac{(1.5\times10^{-3})(3.0\times10^{-4})}{1.5}$

$\Delta \lambda = 3.0\times10^{-7}\,\text{m}$

Convert to nm:

$1\,\text{nm} = 10^{-9}\,\text{m}$

Therefore,

$\Delta \lambda = 3.0\times10^{-7}\times10^9 = 300\,\text{nm}$

The difference in wavelengths of red and violet light is

$\boxed{300}$

nm.