Two coherent plane light waves of equal amplitude makes a small angle $$\alpha\ (<<1)$$ with each other. They fall almost normally on a screen. If $$\lambda$$ is the wavelength of light waves, the fringe width $$\Delta x$$ of interference patterns of the two sets of waves on the screen is

Let the two coherent plane waves be represented by the electric-field phasors

$$\begin{aligned} \mathbf{E}_1 &= E_0 \cos\!\left(\omega t-\mathbf{k}_1\!\cdot\!\mathbf{r}\right),\\ \mathbf{E}_2 &= E_0 \cos\!\left(\omega t-\mathbf{k}_2\!\cdot\!\mathbf{r}\right), \end{aligned}$$
where $$|\mathbf{k}_1| = |\mathbf{k}_2| = k = \dfrac{2\pi}{\lambda}$$.

The waves strike the screen almost normally; their wave-vectors are symmetric about the screen normal and each makes a small angle $$\alpha$$ with that normal. Thus the angle between the two waves is $$2\alpha$$.

Choose the origin on the screen (the screen is the $$z=0$$ plane) and measure distance $$x$$ along the common line of intersection of the two wave fronts with the screen. In this geometry

$$\mathbf{k}_1\cdot\mathbf{r}\Big|_{z=0}=k\,x\sin\alpha,\qquad \mathbf{k}_2\cdot\mathbf{r}\Big|_{z=0}=-k\,x\sin\alpha.$$ Hence on the screen the instantaneous fields are

$$\begin{aligned} E_1(x,t) &= E_0\cos\!\bigl(\omega t-kx\sin\alpha\bigr),\\ E_2(x,t) &= E_0\cos\!\bigl(\omega t+kx\sin\alpha\bigr). \end{aligned}$$

The resultant intensity is proportional to the squared magnitude of the sum:

$$\begin{aligned} I(x) &\propto \bigl[E_1+E_2\bigr]^2 = 4E_0^{\,2}\cos^2\!\bigl(kx\sin\alpha\bigr). \end{aligned}$$

Bright fringes (maxima) occur when the argument of the cosine is an integral multiple of $$\pi$$:

$$k x \sin\alpha = m\pi \quad(m = 0,\,1,\,2,\dots).$$

Therefore the positions of successive bright fringes are

$$x_m = \frac{m\pi}{k\sin\alpha} = \frac{m\lambda}{2\sin\alpha}\;.\tag{1}$$

The fringe width $$\Delta x$$ is the distance between two successive maxima:

$$\Delta x = x_{m+1}-x_m = \frac{\lambda}{2\sin\alpha}.$$

Because $$\alpha$$ is very small, $$\sin\alpha \approx \alpha$$ (in radians). Thus

$$\boxed{\Delta x \approx \dfrac{\lambda}{2\alpha}}.$$

Hence the correct choice is
Option C which is: $$\dfrac{\lambda}{(2\alpha)}$$.