Answer to Question #162511 in Optics for Ulrich

Question

Two slits are separated by 0.320 mm. A beam of 500nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range -30.0degree < s < 30.0degree, where s is an angle

Answer 1

The slit separation is d = 0.320 mm = 0.320$\times$ 10^-3 m

The wavelength of the beam is λ = 500 nm = 500*10^-9 m

Let there be n fringes (bright& dark) of width $x$ on one side of the central bright.

Then the total distance on the screen within the angular separation is Y = $nx$

Then,

$tan\theta= \dfrac{Y}{D}=\dfrac{nx}{D}$

$nx= Dtan\theta$

Where D is the separation of slits and the screen.

The width of each fringe is

$\Rightarrow x=\dfrac{\lambda D}{d}$

$\Rightarrow nx=\dfrac{n\lambda D}{d}$

$\Rightarrow Dtan\theta=\dfrac{n\lambda D}{d}$

$\Rightarrow n= \dfrac{d\times tan\theta}{\lambda}$

$=\dfrac{(0.320\times 10^{-3})(tan30)}{500\times 10^{-9}}$

$=369.5$

The total number fringes in the angular separation of $-30^\degree$to $30^\degree$is 369.5$\times$2 = 739

The total number of bright fringes(including the central bright) is 370