Pressure in the Eye
Pressure in the Eye
The shape of the eye is maintained by fluid pressure, called intraocular pressure, which is normally in the range of 12.0 to 24.0 mm Hg. When the circulation of fluid in the eye is blocked, it can lead to a buildup in pressure, a condition called glaucoma. The net pressure can become as great as 85.0 mm Hg, an abnormally large pressure that can permanently damage the optic nerve. To get an idea of the force involved, suppose the back of the eye has an area of \(6\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}\), and the net pressure is 85.0 mm Hg. Force is given by \(F=\text{PA}\). To get \(F\) in newtons, we convert the area to \({\text{m}}^{2}\) (\({\text{1 m}}^{2}={\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}\)). Then we calculate as follows:
\(F=\mathrm{h\rho }\text{gA}=(\text{85}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m})(\text{13}\text{.}6×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3})(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2})(6\text{.}0×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2})=6.8\text{ N}.\)
Eye Pressure
The shape of the eye is maintained by fluid pressure, called intraocular pressure. When the circulation of fluid in the eye is blocked, it can lead to a buildup in pressure, a condition called glaucoma. The force is calculated as
\(F=\mathrm{h\rho }\text{gA}=(\text{85}\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m})(\text{13}\text{.}6×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3})(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2})(6.0×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2})=6.8\text{ N}.\)
This force is the weight of about a 680-g mass. A mass of 680 g resting on the eye (imagine 1.5 lb resting on your eye) would be sufficient to cause it damage. (A normal force here would be the weight of about 120 g, less than one-quarter of our initial value.)
People over 40 years of age are at greatest risk of developing glaucoma and should have their intraocular pressure tested routinely. Most measurements involve exerting a force on the (anesthetized) eye over some area (a pressure) and observing the eye’s response. A noncontact approach uses a puff of air and a measurement is made of the force needed to indent the eye (see the figure below). If the intraocular pressure is high, the eye will deform less and rebound more vigorously than normal. Excessive intraocular pressures can be detected reliably and sometimes controlled effectively.
The intraocular eye pressure can be read with a tonometer. (credit: DevelopAll at the Wikipedia Project.)
Example: Calculating Gauge Pressure and Depth: Damage to the Eardrum
Suppose a 3.00-N force can rupture an eardrum. (a) If the eardrum has an area of \(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{cm}}^{2}\), calculate the maximum tolerable gauge pressure on the eardrum in newtons per meter squared and convert it to millimeters of mercury. (b) At what depth in freshwater would this person’s eardrum rupture, assuming the gauge pressure in the middle ear is zero?
Strategy for (a)
The pressure can be found directly from its definition since we know the force and area. We are looking for the gauge pressure.
Solution for (a)
\({P}_{\text{g}}=F/A=3\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{N}/(1\text{.}\text{00}×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2})=3\text{.}\text{00}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}.\)
We now need to convert this to units of mm Hg:
\({P}_{\text{g}}=3\text{.}0×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}(\cfrac{\text{ 1.0 mm Hg}}{\text{133}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}})=\text{226 mm Hg.}\)
Strategy for (b)
Here we will use the fact that the water pressure varies linearly with depth \(h\) below the surface.
Solution for (b)
\(P=\mathrm{h\rho g}\) and therefore \(h=P/\mathrm{\rho g}\). Using the value above for \(P\), we have
\(h=\cfrac{3.0×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}}{(1.00×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3})(9.80\text{ m/s}^{2})}=3.06\text{ m}.\)
Discussion
Similarly, increased pressure exerted upon the eardrum from the middle ear can arise when an infection causes a fluid buildup.
This lesson is part of:
Fluid Statics