3 (5 points) image a point source of light using single lens of focal length.The

Question

3 (5 points) image a point source of light using single lens of focal length.The point source radiates within a cone having the full angle + Compute the distances and the angular light spread of the source angular light spread (a) at the image in terms of the image and obiec in degrees and radians d - 60 degrees and the magnification is 3 compute c. 4. (10 points) The power incident on a detector of light is 100 pw a. b. How many Nano-watt is the power level? De termine the number of photons per second incident on the detector if the C. If6.SE 9 photons incident on this detector per second at wavelength 1.3 um.Com d. If this detector converts light to current at the rate of 0.65mA/mw, what current wavelength is 1550 nm. the power incident on the detector produced?

Explanation / Answer

P = 100 pW

a)1 pW = 0.001 nW

100 pW = 0.1 nW

b)We know that,

P = E/t

E = P x t = 100 pW x 1s = 100 pJ = 100 x 10^-12 J

energy of a single photon is:

E = hc/lambda = 6.626 x 10^-34 x 3 x 10^8/(1550 x 10^-9) = 1.28 x 10^-19 J

So number of photon is:

N = 100 x 10^-12/1.28 x 10^-19 =7.81 x 10^8 photons

Hence, N = 7.81 x 10^8 photons

c)we have N = 6.5 x 10^9

lambda = 1.3 um

E = h c/lambda = 6.626 x 10^-34 x 3 x 10^8/(1.3 x 10^-6) = 1.5 x 10^-19 J

No energy of N photons

E = 6.5 x 10^9 x 1.5 x 10^-19 J = 9.75 x 10^-10 J

P = E/t = 9.75 x 10^-10 J/1 s = 9.75 x 10^-10 W

Hence, P = 9.75 x 10^-10 W

d)I/P = 0.65 mA/mW = 0.65 A/W

I = 9.75 x 10^-10 x 0.65 = 6.34 x 10^-10 A

Hence, I = 6.34 x 10^-10 A

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