Question

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6. Light going through a lens will have diffraction effects. A spy satellite in a low earth orbit of 160 km above the surface of the carth uses a lens with a diameter of 45 cm. a. How far apart can two orange (585 nm) basketballs be on the earth and still be resolved? [.25 m) b. What if they were colored purple instead, say 425 nm? Could they be resolved at the distance from a, or would they need to be farther apart to be resolved? Explain. c. For any given color, to get better resolution, should you use a bigger or smaller lens? Explain.

Answer 1

Solution:

The resolving power of the lens is its ability to separate the images of two objects, which are close together.

Each object forms a diffraction pattern. The criteria for resolving is dependent upon the diffraction pattern of the two objects i.e., when it meets the Rayleigh criterion.

Rayleigh criterion says, we are able to resolve two objects as long as the central maxima of the two diffraction patterns do not overlap .

UWA 1.22 2 시미

The condition is

$\theta{min}=\frac{1.22\lambda}{D} =\frac{y}{L}$

$\theta{min}$ is the minimum angular seperation

y is the distance between two objects in meters

L is the distance between object and lens in meters

$\lambda$ is the wavelength in meters

D is the diameter of aperture in meters

Part a - orange:

$\lambda = 585\,nm$

D = 45 cm

L = 160 km

y = unknown.

Solving for y,

$y_{orange}=\frac{1.22\lambda L}{D}= \frac{1.22\times 585 \times 160 \times 10^{-6}}{45\times10^{-2}}=0.253 \,m$

They can be 0.253 meters apart and still be resolved.

Part-b - violet :

we first evaluate the above expression with wavelength of 425 nm

$y_{violet}=\frac{1.22\lambda L}{D}= \frac{1.22\times 425 \times 160 \times 10^{-6}}{45\times10^{-2}}=0.184 \,m$

since this distance is less compared to part-a, they can be resolved even when they are 0.253 m apart and no need to move them apart.

Solution to Part - C:

Bigger the aperture,higher the resolving power. So bigger lens is better.

Resolving power also depends on the distance between the objects and lens. As distance increases, so does the resolving power.