# Angular Fringe Width As θ is very small, then tan θ = y/D can be reduced to

$\large \theta = \frac{y}{D}$

Therefore , $\large \Delta\theta = \frac{\Delta y}{D}$ ..(i)

Where Δy gives the distance between two consecutive similar fringes.

Δy = β , the fringe width.

Hence from (i) we obtain,

$\large \Delta\theta = \frac{\beta}{D}$

But , $\large \beta = \frac{\lambda D}{d}$

Hence , $\large \Delta \theta = \frac{\lambda }{d}$

where Δθ is called the angular fringe width.

Illustration : A beam of light consisting of two wavelengths, 6500 A° and 5200 A° is used to obtain interference fringes in a Young’s double slit experiment (1A° = 10-10 m). The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.

(a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 A°.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Solution : (a) The centre of the nth bright fringe is at a distance

$\large y = \frac{n \lambda D}{d}$ from the central maximum. For the 3rd bright fringe of 6500 A°,

$\large y = \frac{3 \times 6500 \times 10^{-10} \times 1.2}{2 \times 10^{-3}}$

= 0.117 cm ≈ 0.12 cm.

(b) Suppose the mth bright fringe of 6500Å coincides with the nth bright fringe of 5200 A°.

Then, $\large \frac{m \times 6500 A^o D}{d} = \frac{n \times 5200 A^o D}{d}$

$\large \frac{m}{n}= \frac{4}{5}$

The minimum values of m and n that satisfy this equation are 4 and 5 respectively. The distance of the 4th bright fringe of 6500 A°
or the 5th bright fringe of 5200 A° from the central maximum is

$\large y = \frac{4 \times 6500 \times 10^{-10} \times 1.2}{2 \times 10^{-3}}$

= 0.156 cm ≈ 0.16 cm.