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