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Patch of Light

https://isaacphysics.org/questions/light_circ_mirror?stage=a_level

for part d, i get an answer of 17 but isaac physics says im out by a simple factor

i used surface area of cap / surface area of shell = 1/4
which gives me the distance to be 3R/4
and then using trig i made a right angled triangle with R as the hyp, 15 as the adj and 3R/4 as opp and calculated a value for 3R/4 which gives me 17

any help would be appreciated :smile:
Original post by lottte
https://isaacphysics.org/questions/light_circ_mirror?stage=a_level

for part d, i get an answer of 17 but isaac physics says im out by a simple factor

i used surface area of cap / surface area of shell = 1/4
which gives me the distance to be 3R/4
and then using trig i made a right angled triangle with R as the hyp, 15 as the adj and 3R/4 as opp and calculated a value for 3R/4 which gives me 17

any help would be appreciated :smile:


What is the distance (3R/4) that you are referring to?

Your mistake seems to be from the ratio equation:
surface area of cap / surface area of shell = 1/4

It would be better if you show how you write the ratio equation.
Reply 2
Original post by Eimmanuel
What is the distance (3R/4) that you are referring to?

Your mistake seems to be from the ratio equation:
surface area of cap / surface area of shell = 1/4

It would be better if you show how you write the ratio equation.

distance (3R/4) is referring to the distance between the screen and the mirror
i got the equation from using the hints in the video
i did (2*pi*R*h)/(2*pi*R^2) = 1/4
which gives me a value of h=1/4
and since R=x+h,
x = 3R/4
where x is the distance between the screen and mirror
Original post by lottte
distance (3R/4) is referring to the distance between the screen and the mirror
i got the equation from using the hints in the video
i did (2*pi*R*h)/(2*pi*R^2) = 1/4
which gives me a value of h=1/4
and since R=x+h,
x = 3R/4
where x is the distance between the screen and mirror


Note that the question "Assume that as the light source is spherical", the surface area of the shell is NOT 2*pi*R^2 (which is the surface area of the hemisphere).

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