Condition for resolution of two point sources of light

[songoku]

Homework Statement

Please see below

Relevant Equations

$\theta=1.22 \frac{\lambda}{b}$

The formula is $\theta=1.22 \frac{\lambda}{b}$ where b is the diameter of circular aperture

I thought it would be $\theta \geq 0.61 \frac{\lambda}{r}$ since diameter = 2 x radius but the answer is (D)

Do we just consider $r \approx b$ since maybe it is small or am I missing something else?

Thanks

 

[SammyS]

Homework Statement: Please see below
Relevant Equations: $\theta=1.22 \frac{\lambda}{b}$

View attachment 342834

The formula is $\theta=1.22 \frac{\lambda}{b}$ where b is the diameter of circular aperture

I thought it would be $\theta \geq 0.61 \frac{\lambda}{r}$ since diameter = 2 x radius but the answer is (D)

Do we just consider $r \approx b$ since maybe it is small or am I missing something else?

Thanks

I agree with you.

The Rayleigh Criterion says that when two distant lights sources (wavelength, $\lambda\,$) are viewed through a circular aperture of diameter, $D$, they may be resolved provided that there angular separation, $\theta$ (in radians) is such that

$\displaystyle \quad \quad \theta \ge 1.22 \dfrac{\lambda}{D} \ .$

Since $D=2r$ , I agree that the correct answer is C) .

 

[songoku]

Thank you very much SammyS