Question

The distance from the bisector to a point P on the fourth nodal line in a two-point interference pattern is 7.86 cm. The distance between the sources is 0.644 cm. The distance from the midpoint between the two sources to
point P is 18.4 cm. What is the wavelength of the waves?

Answer 1

Solution.

$n=4;$

$x_4=7.86cm;$

$L=18.4cm;$

$d=0.644cm;$

$\lambda-?;$

$\dfrac{x_n}{L}=\dfrac{(n-\dfrac{1}{2})\lambda}{d};$

$\lambda=\dfrac{x_nd}{(n-\dfrac{1}{2})L};$

$\lambda=\dfrac{x_4d}{(n-\dfrac{1}{2})L};$

$\lambda=\dfrac{7.86cm\sdot0.644cm}{(4-\dfrac{1}{2})18.4cm}=0.0786cm=7.86\sdot10^{-2}cm.$

Answer: $\lambda=7.86\sdot10^{-2}cm.$