OphthalWall

[eye see what you did there] 👀 an OD student’s study blog and life in memes

[ophthalwall-blog]

i’m [glad]iolus you came •••

[cwote]

I hope you wake up early up enough in the morning to put on your favorite outfit and make your favorite breakfast. I hope you find your skin clear and hair healthy. I hope you’re content with the way your body looks. I hope you find happiness in small things today. I hope that, even with your ups and downs, you are content with yourself. 

[cozybug-deactivated20231208]

*regularly dreams about owning my own home and turning one of the rooms into a library*

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♬ you’re the cause of my euphoria

[yabookquote]

“A good book will give you answers to questions you didn’t know you had. A great book will give you questions to answers you thought you knew.”

— Beth Revis, Give the Dark My Love

[thenimbus-deactivated20200907]

My jaw dropped when he started talking about the 40 micro arc second angle that the M87 black hole subtends in the sky because I understand what that means now after taking optics courses and HOLY SHIT is that infinitesimal. So I calculated what visual acuity you would need to have to see it, for funsies.

If the black hole is 23.61 billion miles in diameter and your viewing distance is 53.49 million lightyears, the blackhole will subtend 40 micro arc seconds at 0.002 lightyears and, according to my rough calculations in terms of 20/20 snellen acuity, you would need a visual acuity of 20/0.000000000751287 to see it, meaning that you would need around somewhere in the ballpark of 2.67 TRILLION PERCENT better vision than 20/20 to see the M87 black hole.

My work if you want to check me:

If viewing distance is 20 feet (top 20 in 20/20), a 0.35 inch 20/20 sized letter subtends 5’ at 20 feet:

xtan*(2.5’/60˚) = y

x = distance object subtends 5’ of arc (bottom 20 in 20/20)

y = half of the object’s height

If viewing distance is 53.49 million lightyears a 23.61 billion mile tall black hole would subtend 40 μas at 0.002 lightyears:

40 μas/2 = 20 μas = 3.3335x10^-7’

y = 23.61x10^9 miles / 2 = 11.81x10^9 miles

xtan(3.3335x10^-7’/60˚) = 11.81x10^9 miles

x = 1.181x10^10 miles = 0.002 lightyears = 6.23568x10^13 ft

If the black hole subtends 40 μas at 0.002 lightyears and the working distance is 53.49 million lightyears = 3.144x10^20 miles = 1.66x10^24 ft, we solve for the snellen fraction equivalent proportionally:

1.66x10^24 ft/6.23568x10^13 ft = 20ft/x

x = 7.51287x10^-10 ft

7.51287x10^-10 is 2.6621x10^12% smaller than 20

[ophthalwall-blog]

okay, maaaaaaaaybe optics is kinda a little bit cool & I should enjoy it more

[justsomeantifas]

you know reading difficulties are so detrimental to kids in school, as well as adults obvs, and if they aren’t getting proper eye exams they’ll never know whats wrong and whats stopping them from being able to read properly

and you need very detailed eye exams to figure out whats wrong sometimes, and like I personally had access to a nearby university’s school of optometry’s graduate students who are learning, so my insurance covered it because it’s cheaper being that it’s with students

but like most people don’t have access to such detailed eye exams and will never find out why they have so much trouble reading and it’s so upsetting.

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sunday fundays

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hello :)