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  • A mono-chromatic point source S radiating wavelength  with power <img alt="2 \m

A mono-chromatic point source S radiating wavelength \mathrm{6000 \AA} with power 2 \mathrm{w}_1 an aperture A of diameter 0.1 metre and a large screen Sc are placed as shown in figure. A photo emission detector D of surface Area 0.5 \mathrm{~cm}^2 is placed at the centre of the screen. The efficiency of the detector for the photo-electro generation per incident photon is 0.9

Question:

If the concave lens L of focal length 0.6 is inserted in the aperture as shown find the new values of photon current (Assume uniform average transmission is 80% from the lens).

Option: 1

5.33 \times 10^{-8} \mathrm{~A}


Option: 2

2.5 \times 10^{-10} \mathrm{~A}


Option: 3

3.1 \times 10^{-6} \mathrm{~A}


Option: 4

None


Answers (1)

best_answer

Using lens formula

\mathrm{\frac{1}{V}-\frac{1}{-0.6}=-\frac{1}{0.6} \Rightarrow V=-0.3 \mathrm{~m}}

Total number of photon incident per unit time on the lens =

\mathrm{\frac{2}{\left(\frac{\mathrm{hc}}{\lambda}\right)} \times 1.6 \times 10^{-19} \times \frac{1}{4 \pi(0.6)^2} \times \frac{\pi}{4}(\mathrm{~d})^2}

where d = 0.1 meter

\mathrm{=1.052 \times 10^{16} / \mathrm{sec}}

80% of it transmitted to other side. Now source is 5.7 from the detector.

\mathrm{\begin{aligned} & \text { Photo-current }=\frac{1.052 \times 10^{16} \times \frac{80}{100}}{\pi\left(\frac{5.7}{3}\right)^2} \times\left(0.5 \times 10^{-4}\right) \times 0.9 \times 1.6 \times 10^{-19} \\ & =5.33 \times 10^{-8} \mathrm{~A} \end{aligned}}

Posted by

Rishi

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