Q166 / 200 · Manipal (MET) 2014 · Physics

The minimum intensity of light to be detected by human eye is $10^{-10} \mathrm{~W} / \mathrm{m}^2$. The number of photons of wavelength $5.6 \times 10^{-7} \mathrm{~m}$ entering the eye, with pupil area $10^{-6} \mathrm{~m}^2$, per second for vision will be nearly

  • A
    100
  • B
    200
  • Correct answer
    C
    300
  • D
    400

Solution

On using $I=\frac{P}{A}$; where $P=$ radiation power $\Rightarrow \quad P=I \times A \Rightarrow \frac{n h c}{t \lambda}=I A \Rightarrow \frac{n}{t}=\frac{I A \lambda}{h c}$
Hence, number of photons entering per sec the eye $\left(\frac{n}{t}\right)=\frac{10^{-10} \times 10^{-6} \times 5.6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^8}=300$

Hence, the correct answer is option: 'C'.

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