Question

Assuming perfect optics, if the smallest feature you can resolve when observing around 660 nm is of angular size 0.04 arcsec, what would the size (in arcsec) of the finest feature you can resolve if you make your observation around wavelength 1320 nm, assuming everything else is the same

Answer 1

The value is
`\theta _2 = 0.08 \ arcsec`

Step-by-step explanation:

From the question we are told that

The first wavelength is
`\lambda_1 = 660 \ nm = 660 *10^(-9 )`

The first angular size is
`\theta_1 = 0.04 \ arcsec`

The second wavelength is
`\lambda _2 = 1320 \ nm = 1320 *10^(-9 ) \ m`

Generally according to Rayleigh Criterion we have that

`\theta= 1.22 * (\lambda )/(D)`

given every other thing remains constant we have that

`\theta = k * \lambda`

Here k represented constant so

`k = (\theta )/(\lambda)`

=>
`(\theta_1)/( \lambda_1) = (\theta_2)/( \lambda_2)`

=>
`(\theta_1)/( \theta_2) = (\lambda_1)/( \lambda_2)`

So

`( 0.04)/( \theta_2) =0.5`

=>
`\theta _2 = 0.08 \ arcsec`