Question

what should be the width of each slit to obtain n maxima of double slit pattern within the central maxima of single slit pattern?​

Answer 1

Answer:

Explanation:

Let 'a' be the width of each slit.

Linear separation between n bright fringes, x = nβ = \(\frac{n\lambda D}{d}\)

Corresponding angular separation, θ1 = \(\frac{\mathrm{x}}{D}=\frac{n\lambda}{d}\)

Now, the angular width of central maximum in the diffraction pattern of a single slit,

θ2 = \(\frac{2\lambda}{a}\)

As θ1 = θ2

\(\frac{n\lambda}{d}=\frac{2\lambda}{a}\)

\(\therefore\) a = \(\frac{2d}{n}\); where d = separation between slits

OR

Let 'a' be the width of each slit.

Linear separation between n bright fringes, x = nβ = nλDd.

Corresponding angular separation, θ1 = xD=nλd.

θ2 = 2λa.

nλd=2λa.

∴ a = 2dn; where d = separation between slits