Question

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture f/4.00 to photograph a bear that is 12.0m away. Assume the wavelength is 550 nm.
A). What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture?
B).If, to gain depth of field, the photographer stops the lens down to f /22.0, what would be the width of the smallest resolvable feature on the bear?

Answer 1

Answer:

(A) The width $x = 0.24 \times 10^{-3}$ m

(B) The new width is $1.32 \times 10^{-3}$ m

Explanation:

Given :

Focal length $f = 135 \times 10^{-3} m$

Maximum aperture $D = \frac{f}{4}$

Wavelength $\lambda = 550 \times 10^{-9}$ m

(A)

From rayleigh criterion,

  $\theta = \frac{1.22 \lambda }{D}$

  $\theta =\frac{ 1.22 \times 550 \times 10^{-9} }{33.75 \times 10^{-3} }$

  $\theta = 1.98 \times 10^{-5}$ rad

From angle formula,

  $x = R\theta$

Where $R =$ 12 m ( given in example )

$x = 12 \times 1.98 \times 10^{-5}$ m

$x = 23.76 \times 10^{-5}$

$x = 0.24 \times 10^{-3}$ m

(B)

We know that $\theta$ is proportional to the $x$ and inversely proportional to the $D$

so we write the new width, here $x$ is 5.5 times larger than above case

   $x = 0.24 \times 10^{-3} \times \frac{22}{4}$

   $x = 1.32 \times 10^{-3}$ m