Not your usual fringe width

Consider the intensity at a point $(r\cos{\theta} , r\sin{\theta} )..... 0 < \theta < \pi$

The path difference between the light waves reaching the point P can be found using geometry and it comes out to be

$\Delta x = d \sin{\frac{ \theta}{2} }$

( Hint : consider the coordinates of the slits $S_1$ and $S_2$ in polar form and find $| S_{1}P - S_{2}P |$ )

For a Maxima , $\Delta x = n \lambda$ where $\lambda$ is wavelength .

Consider two successive Maximas formed at angles $\theta_1$ , $\theta_2$ .

Since $\dfrac{ \lambda} { d}$ is of the order of a millimeter , we can approximate the distance between two successive Maximas to be

$\beta \approx R ( \theta_2 - \theta_1 ) = R \Delta \theta$

Also we have $\sin{ \frac{\theta_2}{ 2}} - \sin{ \frac{ \theta_1 }{2} } \approx \dfrac{ \cos{ \frac{ \theta_1}{2} } \Delta \theta }{2} = \frac{ \lambda }{ d}$

Using this, we get

$\beta = \dfrac{2 R \lambda }{d \cos{ \frac{ \theta_1 }{2} } }$

When $\theta_1 = 60° , \color{#3D99F6}{ \beta = \dfrac{ 4R \lambda}{d \sqrt{3} } = 0.9237 mm }$