Answer

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**Hint:**To solve this question we have to know about fringes. We know that fringes are the bright or dark bands which are caused by beams of light which are in phase or out of phase.Interference is what happens when two or more waves meet each other.Depending on the alignment of peaks and troughs of the overlapping waves, they might add up, or they can partially or entirely cancel each other.

**Complete step by step answer:**

We can say, the wavelength of the light is \[{\lambda _1} = 650nm\].

Again, the wavelength of second light, \[{\lambda _2} = 520nm\].

Distance between the slit and the screen according to this question is, \[1.4m\].

Distance between the slits is \[0.28nm\].

(a) We know that, the relation between the \[{n^{th}}\] bright fringe and the width of fringe is, \[x = n{\lambda _1}\dfrac{D}{d}\]

Here n is the number of bright fringe

Here the value of n is \[3\]. So,

\[x = 3 \times 650\dfrac{{1.4}}{{0.28 \times {{10}^{ - 3}}}} \\

\Rightarrow x = 1950 \times 5 \times {10^3}nm\]

\[\therefore x = 9.75mm\]

(b) We can now consider that \[{n^{th}}\] bright fringe of \[{\lambda _2}\] and the \[{(n - 1)^{th}}\] bright fringe of wavelength \[{\lambda _1}\] coincide with each other.

\[n{\lambda _2} = (n - 1){\lambda _1}\]

After putting the values

$520n = 650n - 650 \\

\Rightarrow 650 = 130n \\

\Rightarrow n = 5$

Therefore we can say, the least distance from the central maximum can be obtained as.

$x' = n{\lambda _2}\dfrac{D}{d} \\

\Rightarrow x' = 5 \times 520\dfrac{D}{d} = 2600\dfrac{{1.4}}{{0.28 \times {{10}^{ - 3}}}}nm \\ $

\[\Rightarrow x' = 1.30 \times {10^{ - 2}}m \\

\therefore x'= 1.3cm\]

This is the right answer.

**Note:**We know that, for a large wavelength, a large path difference is needed to have a change of phase, then the distance between fringes gets larger. We know that, if the screen is farther for a fixed angle the distance between two fringes gets larger.

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