Respuesta :
Answer: 0.0488 mm
Explanation:
Given
The position of the screen is [tex]D=0.80\ m[/tex]
Distance between alternating bright fringes is [tex]0.95\ cm[/tex]
The wavelength of the light source is [tex]\lambda =580\ nm[/tex]
Distance between successive bright fringes is [tex]\frac{\lambda D}{d}[/tex]
Distance between alternating bright fringes is half of the distance between successive fringes i.e. [tex]\frac{\lambda D}{2d}[/tex]
Insert the values for alternating fringes
[tex]\Rightarrow 0.95\times 10^{-2}=\dfrac{580\times 10^{-9}\times 0.8}{d}\quad [\text{d=separation of the doubleslits}]\\\\\Rightarrow d=\dfrac{464\times 10^{-9}}{0.95\times 10^{-2}}\\\\\Rightarrow d=488.42\times 10^{-7}\ m\\\Rightarrow d=0.0488\ mm[/tex]
