The resolution of a digital cameras is limited by two factors diffraction by the lens, a limit of any optical system, and the fact that the sensor is divided into discrete pixels. consirer a typical point-and--shoot camera that has a $20-mm-focal-length$lens and a sensor with $2.5-\mu m-wide$pixels.

(a) . First, assume an ideal, diffractionless lens, at a distance of $100m,$what is the smallest distance, in $cm$between two point sources of light that the camera can barely resolve? in answering this question, consider what has to happen on the sensor to show two image points rather than one you can use $S^1=fbecauses>>f.$

(b) . You can achieve the pixel-limied resolution of part a only if the diffraction which of each image point no greater than the diffraction width of image point is no greater than $1$pixel in diameter. for what lens diameter is the minimum spot size equal to the width of a pixel ? use $600nm$for the wavelength of light.

(c). what is the $f-number$of the lens for the diameter you found in part b? your answer is a quite realistic value of the $f-number$at which a camera transitions from being pixel limited to being diffraction limited for $f-number$smaller than this (larger-diameter apertures), the resolution is limited by the pixel size and does not change as you change the apertures. for $f-number$larger than this (smaller-diameter apertures). the resolution is limited by diffraction and it gets worse as you "stop down" to smaller apertures.

Step by step solution

01

Part(a) Step 1: Given information

We have given that:

Focal length$f=20mm$,

a diffractionless lens, at a distance$S=100m$,

sensor pixel $h^1=2.5\mu m$and

wavelength $\lambda =600nm$.

We need to find the smallest distance, in cm between two point sources of light that the camera can barely resolve.

02

Part (a) Step 2: Simplification

The magnitude of the lateral magnification is ,

$m=\frac{f}{S}$

Here, $m$is magnification, $f$is focal length and $S$is length with diffractionless lens away.

Substituting the values in the equation,

$m=\frac{20mm}{100m}$

$m=0.00020$

Separating to find $h$,

$h=\frac{h^1}{m}$

$h=\frac{2.5\mu m}{0.00020}$

$h=1.3cm$.

Here,$h$is the smallest distance.

03

part (b) Step 1: Given Information

We need to find the diameter.

04

part (b) step 2: Calculation

For the diameter D,

$\omega =\frac{2.44\lambda L}{D}$

$D=\frac{2.44\lambda L}{\omega }$

Here, $D$is the diameter, $\lambda$is the wavelength and$\omega$is the width of the lens( in pixels )

Substituting the values in the equation,

$D=\frac{2.44(600nm)(20nm)}{2.5\mu m}$

$D=1.2cm.$

05

part (c) step 1: Given Information

We need to find the f-number of the lens for the diameter found in part b.

06

part (c) : Simply

Calculating$f_{number}$,

$f_{number}=\frac{f}{D}=\frac{20}{1.2}\to f_{number}=1.66$.

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