Answer:
[tex]\lambda = 6.25\times10^{-9}[/tex]= 625 nm
Explanation:
We now that for
for maximum intensity(bright fringe) d sinθ=nλ n=0,1,2,....
d= distance between the slits, λ= wavelength of incident ray
for small θ, sinθ≈tanθ= y/D where y is the distance on screen and D is the distance b/w screen and slits.
Given
d=1.19 mm, y=4.97 cm, and, n=10, D=9.47 m
applying formula
λ= (d*y)/(D*n)
putting values we get
[tex]\lambda = \frac{1.19\times10^{-3}\times4.97\times10^{-2}}{9.47\times10}[/tex]
on solving we get
[tex]\lambda = 6.25\times10^{-9}[/tex]= 625 nm