In a biprism experiment , the distance of the `15` th bright band from

1.	In a biprism experiment , the distance of the `15` th bright band from the centre of the interference pattern is 6 mm . Calculate the distance of the 25th bright band and 31st dark band .
Answer» Correct Answer - 10 mm , 12.2 mm Given, 15^th bright β = 6 mm β = \(\frac{\lambda D}d\) 6 mm = \(\frac{15D\lambda}{d}\) \(\frac{\lambda D}d=\frac6{15}\) (i) 25^th bright band β = \(\frac{n \lambda D}d\) β = \(\frac{25\lambda D}d\) (\(\because\) n = 25) \(\because \frac{\lambda D}d=\frac6{15}\) β = \(\frac{25\lambda D}d\) β = 25 x \(\frac6{15}\) β = 30/3 β = 10 mm fringe width for dark band β = (2n + 1)\(\frac{\lambda D}d\) fringe = (2 x 31 + 1) \(\frac{\lambda D}d\) β = 63\(\frac{\lambda D}d\) β = 63 x \(\frac6{15}\) β = 25.2 mm

In a biprism experiment , the distance of the `15` th bright band from the centre of the interference pattern is 6 mm . Calculate the distance of the 25th bright band and 31st dark band .

Answer» Correct Answer - 10 mm , 12.2 mm

Given, 15^th bright β = 6 mm

β = \(\frac{\lambda D}d\)

6 mm = \(\frac{15D\lambda}{d}\)

\(\frac{\lambda D}d=\frac6{15}\)

(i) 25^th bright band

β = \(\frac{n \lambda D}d\)

β = \(\frac{25\lambda D}d\) (\(\because\) n = 25)

\(\because \frac{\lambda D}d=\frac6{15}\)

β = \(\frac{25\lambda D}d\)

β = 25 x \(\frac6{15}\)

β = 30/3

β = 10 mm

fringe width for dark band

β = (2n + 1)\(\frac{\lambda D}d\)

fringe = (2 x 31 + 1) \(\frac{\lambda D}d\)

β = 63\(\frac{\lambda D}d\)

β = 63 x \(\frac6{15}\)

β = 25.2 mm

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