Answer:
The angular width is "[tex]1.575\times 10^{-3} \ rad[/tex]" and the linear width is "[tex]2.835\times 10^{-3} \ m[/tex]".
Explanation:
Given:
Wavelength,
[tex]\lambda = 630 \ nm[/tex]
or,
[tex]=630\times 10^{-9} \ m[/tex]
Slit width,
[tex]a = 0.8 \ mm[/tex]
or,
[tex]=0.8\times 10^{-3} \ m[/tex]
Distance between slit and screen,
[tex]D = 1.8 \ m[/tex]
As we know,
The central bright fringe's angular height,
⇒ [tex]\theta = \frac{2 \lambda}{a}[/tex]
[tex]= \frac{2\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]
[tex]=1.575\times 10^{-3} \ rad[/tex]
and,
The linear width will be:
⇒ [tex]x_o=\frac{2D \lambda}{a}[/tex]
By substituting the values, we get
[tex]=\frac{3\times 1.8\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]
[tex]=2.835\times 10^{-3} \ m[/tex]
