Question

What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two slits is 1 mm?

Answer 1

Let us assume that width of each slit = a.

In a double slit experiment, the separation between Maxima (m) is given as

y = m(Lambda) D/d

D = distance between screen and slit

d = Separation between the slits.

The angular separation between m Maxima is described as

Ф = y/D = m(Lambda) D/(D*d) = m(Lambda)/d

We can derive the angular separation between 10 bright fringes by substituting m = 10

Theta = 10(Lambda)/d = 10(Lambda)/1 = 10(Lambda)  (d = separation between the slits = 1 mm)

Angular width of the central maximum in the single slit differential pattern of width ‘a’ is described as  

2(Ф1) = 2(Lambda)/a

From the question, it is clear that 10 maxima of the double slit pattern is formed within the central maximum of the single slit pattern.

Hence above 2 equations can be equated.

10(Lambda) = 2(Lambda)/a

∴ a = 2/10 = 0.2mm.

The width of each slit = 0.2mm.

Answer 2

Answer:

Let width of a single slit = a

Distance between slits = d = 1mm

In double slit experiment,

Separation between n maxima is given by,

$y_{n} = n \frac{\lambda D}{d}$

Angular separation between n maxima can be given as,

$\theta_{n} = \frac{n \lambda }{d}$

So, angular separation between 10 maxima, $\theta_{10} = \frac{10 \lambda }{d}$

Path difference = asin$\theta$ ≈ a$\theta$= $\lambda$

$=> \theta = \frac{ \lambda }{a}$

The angular width of the central maximum in the diffraction pattern due to single slit of width ‘a’ is given by,

$2\theta = 2(\frac{ \lambda }{a})$

From given data:

$=> \frac{10 \lambda }{d} = \frac{2 \lambda }{a}$

$=> \frac{5}{d} = \frac{1}{a}$

$=> a = \frac{d}{5}$

$=> a = \frac{1}{5} mm$

$=> a = 0.2 mm$

So, width of each slit is 0.2 mm