# Optics physics intensity question

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#1
Part D
https://isaacphysics.org/questions/l...8-46bf75ff7f05

I’m trying to make the area of the cap/ area of sphere =0.25 but I keep getting equations involving h and x (obviously I’m trying to find x so that’s fine) but I can’t get rid of x. I found the video in the final hint helpful so far but could do with a little extra nudge
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7 months ago
#2
(Original post by Jsmithx)
Part D
https://isaacphysics.org/questions/l...8-46bf75ff7f05
I’m trying to make the area of the cap/ area of sphere =0.25 but I keep getting equations involving h and x (obviously I’m trying to find x so that’s fine) but I can’t get rid of x. I found the video in the final hint helpful so far but could do with a little extra nudge
I am not sure are you seeking help or … Because you said that “I found the video in the final hint helpful so far but could do with a little extra nudge” So you have given yourself the extra nudge or you are asking others to give you the extra nudge.

Next, your explanation is also weird. “I’m trying to make the area of the cap/ area of sphere =0.25 but I keep getting equations involving h and x (obviously I’m trying to find x so that’s fine) but I can’t get rid of x.” You want to find x but at the same time, you want to get rid of x. So what do you really want to do? I have never seen a method in maths that you can get rid of x and then solve for x.
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#3
(Original post by Eimmanuel)
I am not sure are you seeking help or … Because you said that “I found the video in the final hint helpful so far but could do with a little extra nudge” So you have given yourself the extra nudge or you are asking others to give you the extra nudge.

Next, your explanation is also weird. “I’m trying to make the area of the cap/ area of sphere =0.25 but I keep getting equations involving h and x (obviously I’m trying to find x so that’s fine) but I can’t get rid of x.” You want to find x but at the same time, you want to get rid of x. So what do you really want to do? I have never seen a method in maths that you can get rid of x and then solve for x.
I’m looking for a nudge (I’m nearly there but not quite)
The x thing is a typo I’m trying to eliminate h (sorry)
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7 months ago
#4
(Original post by Jsmithx)
I’m looking for a nudge (I’m nearly there but not quite)
The x thing is a typo I’m trying to eliminate h (sorry)
It is again best that you post your working.
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#5
2πh(x^2 + 0.3^2)^1/2 (area of cap) divided by 4/3π(x+h)^3 (area of sphere) = 1/4
4h(x^2 + 0.09)^1/2 = 1/3 (x+h)^3
144h^2 (x^2 + 0.09)= (x+h)^6
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7 months ago
#6
(Original post by Jsmithx)
2πh(x^2 + 0.3^2)^1/2 (area of cap) divided by 4/3π(x+h)^3 (area of sphere) = 1/4
4h(x^2 + 0.09)^1/2 = 1/3 (x+h)^3
144h^2 (x^2 + 0.09)= (x+h)^6
4/3π(x+h)^3 is the formula for volume instead of the surface area of a sphere. 0
#7
(Original post by Eimmanuel)
4/3π(x+h)^3 is the formula for volume instead of the surface area of a sphere. I think by looking at the last hint R=x+h
2πh(x+h)/2π(x+h)^2 = 1/4
4h=x+h
x=3h which seems reasonable but I still don’t know what h is
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6 months ago
#8
(Original post by Jsmithx)
I think by looking at the last hint R=x+h
2πh(x+h)/2π(x+h)^2 = 1/4
4h=x+h
x=3h which seems reasonable but I still don’t know what h is
The following quote is incorrect.
2πh(x+h)/2π(x+h)^2 = 1/4
2π(x+h)^2 is the surface area of a hemisphere and this leads to an incorrect relationship between x and h (x=3h).

Note that there is a right-angled relationship between R, h and d.
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#9
(Original post by Eimmanuel)
The following quote is incorrect.

2π(x+h)^2 is the surface area of a hemisphere and this leads to an incorrect relationship between x and h (x=3h).

Note that there is a right-angled relationship between R, h and d.
I got it! Thank you
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