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 (in nm).

We know that in Young’s double-slit experiment the position of the $$m^{\text{th}}$$ bright fringe from the central white fringe is given by the formula

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

where

$$$y_m=\text{distance of the }m^{\text{th}}\text{ bright fringe from the centre},$$$

$$\lambda=\text{wavelength of light},$$

$$$D=\text{distance between the slits and the screen},$$$

$$$d=\text{separation between the two slits}.$$$

Here the first bright fringe for any colour corresponds to $$m=1$$, so for violet light we have

$$y_v=\frac{\lambda_v D}{d}$$

and for red light

$$y_r=\frac{\lambda_r D}{d}.$$

We are supplied with the following numerical data:

$$$D=1.5\ \text{m},\qquad d=0.3\ \text{mm}=0.3\times10^{-3}\ \text{m}=3.0\times10^{-4}\ \text{m},$$$

$$$y_v=2.0\ \text{mm}=2.0\times10^{-3}\ \text{m},\qquad y_r=3.5\ \text{mm}=3.5\times10^{-3}\ \text{m}.$$$

First we find the wavelengths individually. For violet light, substituting the known values gives

$$$\lambda_v=\frac{y_v d}{D}=\frac{2.0\times10^{-3}\,\text{m}\,\times\,3.0\times10^{-4}\,\text{m}}{1.5\ \text{m}}.$$$

Multiplying the numerators,

$$2.0\times10^{-3}\times3.0\times10^{-4}=6.0\times10^{-7},$$

and then dividing by $$1.5$$,

$$$\lambda_v=\frac{6.0\times10^{-7}}{1.5}=4.0\times10^{-7}\ \text{m}=400\ \text{nm}.$$$

In an exactly similar manner for red light we have

$$$\lambda_r=\frac{y_r d}{D}=\frac{3.5\times10^{-3}\,\text{m}\,\times\,3.0\times10^{-4}\,\text{m}}{1.5\ \text{m}}.$$$

Again multiplying the numerators,

$$$3.5\times10^{-3}\times3.0\times10^{-4}=10.5\times10^{-7}=1.05\times10^{-6},$$$

and dividing by $$1.5$$,

$$$\lambda_r=\frac{1.05\times10^{-6}}{1.5}=7.0\times10^{-7}\ \text{m}=700\ \text{nm}.$$$

We are asked for the difference in wavelengths, so we subtract:

$$\Delta\lambda=\lambda_r-\lambda_v.$$

Putting in the two values just obtained,

$$\Delta\lambda=700\ \text{nm}-400\ \text{nm}=300\ \text{nm}.$$

So, the answer is $$300\ \text{nm}.$$