Physics help In this question, model the eye as a single lens (n = 1.4) with air
ID: 1416360 • Letter: P
Question
Physics help
In this question, model the eye as a single lens (n = 1.4) with air in front of it and fluid (n = 1.3) behind it. The eyeball is spherical, of diameter 2.5 cm. The front surface of the lens has the same curvature as the eyeball. In order to model the eye focusing on a distant object, assume that the power of the lens in this system is 52 D. While talking to someone, with your face 20 cm away from theirs, you notice you can see your own reflection in the other person's eyes. The image is right way up. What is the vergence of light scattered off your face as it is about to strike the other person's eyes? What is the reflective power of the front surface of the eye? Characterise the image formed by reflection as either real or virtual. Calculate the image location and magnification of the image formed by reflection, and compare with your expectations of the image. A laser pointer produces a circular spot of red light that barely changes in size as the beam propagates. It is labelled "lambda = 650 nm. Output = 1.5 mW", where 1 mW = 1 mJ/s. Use a ray diagram and imaging calculation to explain what would happen to the laser beam if it entered the eye described above. If laser light hit the eye, the blink reflex would be activated. Assume that the time taken for a person to react and blink varies in the range 0.20 plusminus 0.06 s. Calculate, including uncertainty, the number of photons that would reach the eye before it closed. Laser safety standards use the concept of Maximum Permissible Exposure (MPE), which is the concentration of energy that has a negligible probability to cause damage. For the laser as described here, the MPE is 32 J m^-2 at the front surface of the eye. Assuming that the laser energy is distributed evenly across a 1.8 mm radius beam, would you consider this laser safe for the person in ) (ii)? Justify your answer.Explanation / Answer
a) i) vergence of light = 1.3/52 = 0.025
ii) reflective power = (2.5 * 20)/(2.5 + 20) = 2.22 D
iii) Image formed is virtual .
iv) Here, 1/v = 1/2.5 - 1/20
=> image location , v = 2.857 cm towards right
=> magnification = - 2.857/20 = - 0.1428
b) i) Laser beam would form a very diminished circular spot of red light .
ii) number of photons = (1.5 * 10-3 * 0.20 )/(3.06 * 10-19)
= 9.804 * 1014
iii) Here, energy = 32 * 4 * 3.14 * 1.8 * 10-3 * 1.8 * 10-3
= 1.302 mJ
=> this laser is not safe for person due to its higher energy .