Question

If two point sources of light are being imaged by this telescope, what is the maximum wavelength λ at which the two can be resolved if their angular separation is 3.0×10−5 radians?

Answer 1

Answer:

The maximum wavelength is 492 nm.

Explanation:

Given that,

Angular separation $\theta=3.0\times10^{-5}\ rad$

Suppose a telescope with a small circular aperture of diameter 2.0 cm.

We need to calculate the maximum wavelength

Using formula of angular separation

$\sin\theta=\dfrac{1.22\lambda}{d}$

$\lambda=\dfrac{d\sin\theta}{1.22}$

Put the value into the formula

$\lambda=\dfrac{2.0\times\sin(3\times10^{-5})}{1.22}$

For small angle $\sin\theta\approx\theta$

$\lambda=\dfrac{0.02\times3\times10^{-5}}{1.22}$

$\lambda=4.92\times10^{-7}\ m$

$\lambda=492\ nm$

Hence, The maximum wavelength is 492 nm.

Answer 2

The maximum wavelength ( λ ) at which two can be resolved is : 492 nm

Given data :

Angular separation = 3 * 10⁻⁵ radians

Determine the maximum wavelength

let's assume the telescope is a small circular aperture

Diameter = 2.0 cm = 0.02 m

Applying the formula of angular separation

Sin ∅ = ( 1.22 λ ) / d

Therefore :

λ = dsin ∅ / 1.22  ----- ( 1 )

where : d = 0.02m , ∅ = 3 * 10⁻⁵

Insert values into equation ( 1 )

λ = 0.02 * sin (3 * 10⁻⁵) / 1.22

  =  492 nm

Therefore we can conclude that The maximum wavelength ( λ ) at which two can be resolved is : 492 nm

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