Answer:
[tex]0.0047^{\circ}[/tex]
[tex]0.018978^{\circ}[/tex]
B) Dim
Explanation:
[tex]\theta[/tex] = Angle
a = Pupil diameter
[tex]\lambda[/tex] = Wavelength = 543 nm
Angular resolution is given by
[tex]sin\theta=1.22\dfrac{\lambda}{a}\\\Rightarrow \theta=sin^{-1}\left(1.22\dfrac{\lambda}{a}\right)\\\Rightarrow \theta=sin^{-1}\left(1.22\times \dfrac{543\times 10^{-9}}{8\times 10^{-3}}\right)\\\Rightarrow \theta=0.0047^{\circ}[/tex]
For the dim light the angle is [tex]0.0047^{\circ}[/tex]
[tex]sin\theta=1.22\frac{\lambda}{a}\\\Rightarrow \theta=sin^{-1}\left(1.22\dfrac{\lambda}{a}\right)\\\Rightarrow \theta=sin^{-1}\left(1.22\times \dfrac{543\times 10^{-9}}{2\times 10^{-3}}\right)\\\Rightarrow \theta=0.018978^{\circ}[/tex]
For the bright light the angle is [tex]0.018978^{\circ}[/tex]
The angular separation and sharpness are inversely related. Here, the dim light will produce a sharp image.
Hence, option B is correct
