Answer:
0.000003782 m
0.000001891 m
0.000001197125 m
Explanation:
[tex]\lambda[/tex] = Wavelength = 248 nm
D = Diameter of beam = 1 cm
f = Focal length = 0.625 cm
The angle is given by
[tex]\theta=\dfrac{1.22\lambda}{D}[/tex]
The width is given by
[tex]d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m[/tex]
The required width is 0.000003782 m
Minimum resolvable line separation is given by
[tex]\dfrac{0.000003782}{2}=0.000001891\ m[/tex]
The minimum resolvable line separation between adjacent lines is 0.000001891 m
when [tex]\lambda=157\ nm[/tex]
[tex]d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m[/tex]
The new minimum resolvable line separation between adjacent lines is
[tex]\dfrac{0.00000239425}{2}=0.000001197125\ m[/tex]