Question 280: Wave Optics

Question

In biprism experiment, the fringe width is 0.6 mm . The distance between $6^{	ext {th }}$ dark fringe and $8^{	ext {th }}$ bright fringe on the same side of central bright fringe is

Accepted Answer

In the biprism experiment, we are given that the fringe width is 0.6 mm. We need to determine the distance between the 6th dark fringe and the 8th bright fringe on the same side of the central bright fringe.
To find this distance, let's consider:
Position of the 8th Bright Fringe:
$    x_8 = 8X    $
Position of the 6th Dark Fringe:
$    x_6 = \left(6 - \frac{1}{2}ight) X = 5.5X    $
Calculating the Distance Between the 6th Dark Fringe and the 8th Bright Fringe:
$    x_8 - x_6 = 8X - 5.5X = 2.5X    $
Since the fringe width ($X$) is 0.6 mm, substitute this value to find the distance:
$ x_8 - x_6 = 2.5 	imes 0.6 = 1.5 \, 	ext{mm} $
Therefore, the distance between the 6th dark fringe and the 8th bright fringe is 1.5 mm.

Answer

6 mm

Answer

4 mm

Answer

1.5 mm

Answer

0.9 mm

Wave Optics

In biprism experiment, the fringe width is 0.6 mm . The distance between 6^{\text {th }} dark fringe and 8^{\text {th }} bright fringe on the same side of central bright fringe is

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