Q15s- A power density of 50 W/cm2 can be absorbed by the rods and the cones foun

Question

Q15s- A power density of 50 W/cm2 can be absorbed by the rods and the cones found in the retina beyond this power density, thermal damage occurring in the retina a dye laser is tuned at   = 47 1nm and is impinging on the cornea then focused on the retina. The focusing system of the eye has a focal length of about it; find the power density on the cornea of the eye. W0 near the eye is 1 mm and the laser is 5 mW?

A. 1.59 x10-3 W/cm2

B. 1.59 W/cm2

C. 159 W/cm2

D).0.159 W/cm2

I = 5 mW = 5 x 10^-3 W

We know tat I = P/A

I = 5 x 10^-3 W / 3.14 x (0.001)^2 = 159 W/m^2

I = 159 W/m^2 = 0.159 W/cm^2

Q16- Use the data in questions 15s and calculate the laser power at which thermal damage startsoccurring inretina?

a)       A.5.64 x 10-6W

b)       B.1.41 x10-6W

c)       C.4.72 x 10-2W

d)       D.9.42 x 10-2W

e)       E.none of the above

Explanation / Answer

Q15. given that a power density of 50 W/cm^2 or above is harmful to retina of the eye
given, lambda = 471 nm
area of the region in the eye, A = pi*(0.001)^2
so intensity, I = power/area
power = 5 mW
so intensity = 5*10^-3/pi(0.001)^2 = 1591.5494 W/m^2 = 0.159154 W/cm^2
option D is the answer

Q16. let the damaging laser power be P
then P/A = 50 W/cm^2 = 50*10^4 W/m^2

so, P = 50*10^4*pi*(0.001)^2 = 1.5707 W
so, option E) none of these is the answer

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