Answer:
Distance between slits, [tex]d=2.89\times 10^{-7}\ m[/tex]
Explanation:
It is given that,
Wavelength, [tex]\lambda=410\ nm=410\times 10^{-9}\ m[/tex]
Angle, [tex]\theta=45[/tex]
We need to find the distance between two slits that produces first minimum. The equation for the destructive interference is given by :
[tex]d\ sin\theta=(n+\dfrac{1}{2})\lambda[/tex]
For first minimum, n = 0
So, [tex]d\ sin\theta=(\dfrac{1}{2})\lambda[/tex]
d is the distance between slits
So, [tex]d=\dfrac{1/2\lambda}{sin\theta}[/tex]
[tex]d=\dfrac{1/2\times 410\times 10^{-9}}{sin(45)}[/tex]
[tex]d=2.89\times 10^{-7}\ m[/tex]
So, the distance between two slits is [tex]2.89\times 10^{-7}\ m[/tex]. Hence, this is the required solution.
