Question 294: Wave Optics

Question

The fringe width in an interference pattern is ' X '. The distance between the sixth dark fringe from one side of central bright band to the fourth bright fringe on other side is

Accepted Answer

The fringe width in an interference pattern is denoted by $ X $.
Explanation:
The fringe width is defined as:
$ W = \frac{\lambda D}{d} = X $
Here, $ \lambda $ is the wavelength, $ D $ is the distance to the screen, and $ d $ is the slit separation.
Position of the 4th Bright Fringe:
The position of the $ n $-th bright fringe is given by:
$ n \frac{\lambda D}{d} = 4X $
In this case, for the 4th bright fringe, $ n = 4 $. Thus, the position is $ 4X $.
Position of the 6th Dark Fringe:
The position of the $ n $-th dark fringe is given by:
$ \left(2n - 1ight) \frac{\lambda D}{2d} $
For the 6th dark fringe, $ n = 6 $:
$ (2 	imes 6 - 1) \frac{X}{2} = 5.5X $
Total Distance:
The total distance from the 6th dark fringe on one side of the central bright band to the 4th bright fringe on the other side is:
$ (4 + 5.5)X = 9.5X $
Therefore, the total distance between the specified fringes is $ 9.5X $.

Answer

1.5 X

Answer

2 X

Answer

5.5 X

Answer

9.5 X

Wave Optics

The fringe width in an interference pattern is ' X '. The distance between the sixth dark fringe from one side of central bright band to the fourth bright fringe on other side is

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