Question

Light of wavelength 440nm in air falls on two slits 7.00x 10^-2mm apart. The slits are immersed in water, as is a viewing screen 60.0cm away.

How far apart are the fringes on the screen in meters?

Answer 1

The wavelength of light in water is,

$$ \lambda_{\mathrm{n}}=\frac{\lambda}{n} $$

In double slit experiment, the condition for maxima is

$$ \begin{array}{l} d \sin \theta=m \lambda_{\mathrm{n}} \\ d\left(\frac{y}{D}\right)=m \lambda_{\mathrm{n}} \\ y=\frac{m \lambda_{\mathrm{n}} D}{d} \end{array} $$

The condition for next maxima,

\(d \sin \theta=(m+1) \lambda_{\mathrm{n}}\)

$$ \begin{array}{c} d\left(\frac{y_{1}}{D}\right)=(m+1) \lambda_{\mathrm{n}} \\ y_{1}=\frac{(m+1) \lambda_{\mathrm{n}} D}{d} \end{array} $$

The fringe width is

$$ \begin{aligned} \Delta y &=y_{1}-y \\ &=\frac{(m+1) \lambda_{\mathrm{n}} D}{d}-\frac{m \lambda_{\mathrm{n}} D}{d} \\ &=\frac{\lambda_{\mathrm{n}} D}{d} \\ &=\frac{\lambda D}{n d} \\ &=\frac{\left(400 \times 10^{-9} \mathrm{~m}\right)(0.60 \mathrm{~m})}{(1.33)\left(7.00 \times 10^{-5} \mathrm{~m}\right)} \\ &=2.58 \times 10^{-3} \mathrm{~m} \end{aligned} $$