Question

Between this and the next assignment, we want to get a better under- standing of how light interacts with the eye. Here are two questions to get us started, focused on diffraction (i.e., the spreading of light when it passes through a narrow opening) A. To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright? You must have noticed that when you squint, blurry objects suddenly become a bit clearer. Based on what you found, why does squinting help? B. The pupil of a cat's eye is round under low light, but becomes a thin vertical slit (0.5 mm wide) in bright light (that's thanks to special muscles to narrow it). What are the three smallest angles on either side of the central maximum at which no light will reach the cat's retina? Would the cat be aware of the pattern of alternating dark and light fringes? A few things you can assume: (1) the bright light has a wavelength of 550 nm in air; (2) light enters the eye perpendicular to the lens; and (3) the index of refraction of eye fluid is approximately 1.4.

Answer 1

Considering diffraction through a circular aperture, then angular resolution is given by

${\displaystyle \theta =1.220{\frac {\lambda }{D}}}$

where θ is the angular resolution (radians), λ is the wavelength of light, and D is the diameter of the lens' aperture.

For D = 2mm = 2 x 10⁶ nm

550_ 3.35 × 10-4 radians Θ 1.220 ×

1 arcsecond = 4.85 x 10^-6 radians

So, 3.35 x10^-4 radians =

69.2arcsecnds

Now for D = 8mm

550 8.387 × 10-)radians Θ = 1.220 × S 106

e17.3arcsec

Sharpness of image increases with increasing angular resolution. So, Objects are sharp in dim light.

When something is not clear, it usually means the light being focused by your lens is not focused properly. When you squint, you are reducing the amount of light coming from other sources, this prevents the unfocused light rays in the periphery from reaching the retina and hence only a small amount of focused central light rays are allowed into the eye. The result is better clear image. It is same effect as looking through a pinhole.

B)For destructive interference in a single slit(dark fringe),

$D\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda },\phantom{\rule{0.25em}{0ex}}\text{for}\phantom{\rule{0.25em}{0ex}}m=\text{1,}\phantom{\rule{0.25em}{0ex}}\text{-1,}\phantom{\rule{0.25em}{0ex}}\text{2,}\phantom{\rule{0.25em}{0ex}}\text{-2,}\phantom{\rule{0.25em}{0ex}}\text{3,}\phantom{\rule{0.25em}{0ex}}\dots \phantom{\rule{0.25em}{0ex}}\text{(destructive),}$

where = slit width, $\lambda$ = light’s wavelength, $\theta$ = angle relative to the original direction of the light, and = order of the minimum.

For m= 1

$\Theta = Sin^{-1}(\frac{m\lambda }{D})$

550 0.5 x 106 Sin-1(

       = 0.06^o

For m =2

550 0.5 x 160.126 0.5× 100

For m =3

550 Sin-1 (3 × ) = 0.1890

This difference is very low , so Cat would not be aware of dark and bright pattern