Question

Angle for 3rd maximum when wavelength 580nm and slit seperation 0.1mm

Answer 1

Solution

According to bragg equation

$2d\sin\theta= n\lambda$

Here given n=3

$\lambda=580 \times 10^{-9}m$

d=0.0001m

So angle of third maximum

$\sin \theta=\frac{3\times580 \times 10^{-9}}{2\times 0.0001}$

$\sin \theta=0.00870$

$\theta=0.50^°$