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Find the distance in nm between two slits that produces the first minimum for 405-nm violet light at an angle of 57. 5°

Sagot :

The distance between two slits is d =2.89*10^-7 m

Distance between slits,   d=2.89*10^-7 m

It is given that,

Wavelength, λ = 410nm= 410*10^-9 m

Angle, θ =45

We need to find the distance between two slits that produces first minimum. The equation for the destructive interference is given by :

dsinθ =(n+1/2) λ

For first minimum, n = 0

dsinθ =(1/2) λ

So, d is the distance between slits

d ={1/2 λ}sinθ

=2.89*10^-7 m

So, the distance between two slits is d =2.89*10^-7 m. Hence, this is the required solution.

For more information about slits, visit

https://brainly.com/question/15585828

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