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Chapter 36: Problem 42

A wildlife photographer uses a moderate telephoto lens of focal length \(135 \mathrm{~mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is \(11.5 \mathrm{~m}\) away. Assume the wavelength is \(550 \mathrm{nm}\). (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?

Short Answer

Expert verified
The width of the smallest feature that can be resolved when the lens is opened to its maximum aperture is approximately \(3.24 \times 10^{-4}\) m or 0.324 mm. When the lens is stopped down to f/22, the width of the smallest resolvable feature is approximately \(7.16 \times 10^{-4}\) m or 0.716 mm.

Step by step solution

01

Convert all units to meters

The distance to the bear is given as 11.5 m, the focal length is given as 135 mm = 0.135 m, and the wavelength is given as 550 nm = \(5.50 \times 10^{-7}\) m. We will use these values in meters.
02

Calculate the size of features resolvable at maximum aperture

Substitute the given values into the formula: \(d = 1.22 \times 5.50 \times 10^{-7} \frac{11.5}{4}\). This will give us the smallest resolvable feature when the lens is opened to its maximum aperture.
03

Calculate the size of features resolvable at f/22

Substitute the given values into the formula again, but this time use f=22 instead of f=4: \(d = 1.22 \times 5.50 \times 10^{-7} m \frac{11.5 m}{22}\). This will give us the smallest resolvable feature when the lens is stopped down to f/22.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffraction-Limited Resolution
In the fascinating world of optics, the term diffraction-limited resolution refers to the finest detail that a lens or optical system can resolve. This limit is influenced by the wave nature of light. As light passes through an aperture, such as a camera lens, it spreads out or diffracts, making it challenging to distinguish between two closely positioned points.

The practical implication of this phenomenon for photographers, like the wildlife photographer in our exercise, is that there is a minimum size of the object's detail that can be captured, regardless of how perfect the lens is. This is why even with a high-quality telephoto lens, the wildlife photographer cannot resolve arbitrarily small features on a bear from a distance.

Improving the diffraction-limited resolution can often involve using a lens with a larger aperture, which allows more light to enter and thus can decrease the effect of diffraction, up to a certain point. This principle is used in the solution to calculate the smallest feature on the bear that can be resolved at maximum aperture versus a smaller aperture.
Rayleigh Criterion
One might wonder how we define the 'smallest feature' that can be resolved. This is where the Rayleigh criterion comes into play. It is a critical concept in optical resolution that determines when two point light sources, like stars in a telescope's view or details on a bear's fur, can be seen as distinct. According to this criterion, two features are considered to be resolvable if the central peak of the diffraction pattern of one feature coincides with the first minimum of the other鈥檚 diffraction pattern.

The equation stemming from the Rayleigh criterion involves a factor of 1.22 (which is derived from calculations of a circular aperture's diffraction pattern), the wavelength of light, and the ratio of the focal length to the lens diameter (known as the f-number). This equation, which was used to calculate the smallest feature on the bear in our example, highlights the interplay between the lens specifications and the physics of light.
Focal Length
To understand how a wildlife photographer captures crisp images, we need to consider the focal length of the lens. The focal length, measured in millimeters, determines the lens's angle of view and, thus, its magnification. Typically, the larger the focal length, the higher the magnification, allowing distant objects to be captured in greater detail.

In the case of our bear photographer, a 135mm telephoto lens is used, which has a moderate angle of view and is excellent for isolating subjects from their background. When the exercise considers the impact of changing the aperture size, the focal length remains unchanged. This consistency is crucial as it affects not only the composition of the photograph but also the diffraction-limited resolution. The focal length's relationship with the aperture size (the f-number) is direct; as the photographer stops down the lens from f/4 to f/22, the lens's opening becomes smaller, directly affecting the size of the smallest resolvable feature according to the calculations provided.

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Most popular questions from this chapter

Light of wavelength \(585 \mathrm{nm}\) falls on a slit \(0.0666 \mathrm{~mm}\) wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest that \(m\) can be? \((\mathrm{b})\) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Monochromatic light from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide. On a screen \(2.00 \mathrm{~m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be \(1.35 \mathrm{~mm}\). Calculate the wavelength of the light.

An interference pattern is produced by light of wavelength \(580 \mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of \(0.530 \mathrm{~mm} .\) (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit interference maxima? (b) Let the slits have width \(0.320 \mathrm{~mm} .\) In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

(a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time \(t,\) the electric field at a distant point \(P\) is $$ \begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned} $$ where \(E_{0}\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrangement), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\). (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship \(e^{i z}=\cos z+i \sin z,\) where \(i=\sqrt{-1} .\) In this expression, \(\cos z\) is the real part of the complex number \(e^{i z}\), and \(\sin z\) is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity $$ \sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)} $$ (c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A},\) show that the sum in part \((b)\) can be written as $$ \begin{aligned} E_{0} &\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)} \\\ &=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{j[k R-\omega t+(N-1) \phi / 2]} \end{aligned} $$ Then, using the relationship \(e^{i z}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$ E_{P}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2] $$ The quantity in the first square brackets in this expression is the amplitude of the electric field at \(P\). (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$ I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2} $$ where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N=2\). It will help to recall that \(\sin 2 A=2 \sin A \cos A .\) Explain why your result differs from Eq. (35.10) the expression for the intensity in two-source interference, by a factor of \(4 .\) (Hint: Is \(I_{0}\) defined in the same way in both expressions?)

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is \(656.45 \mathrm{nm} ;\) for deuterium, the corresponding wavelength is \(656.27 \mathrm{nm}\). (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.
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