Question

The minimum light intensity that can be perceived by eye is about $10^{-10} W m ^{-2}$. The number of photons of wavelength $5.6 \times 10^{-7} m$ that must enter the pupil of area $10^{-6} m ^{2}$ per second for vision is approximately (Take $\left.h=6.6 \times 10^{-34} J s ^{-1}\right)$

• A. $3 \times 10^{2}$
• B. $3 \times 10^{3}$
• C. $3 \times 10^{4}$
• D. $3 \times 10^{5}$

Answer 1

Answer: (A)

Here, $I=10^{-10} W m ^{-2}=10^{-10} J s ^{-1} m ^{-2}$
Let the number of photons that enter the pupil per second be
$N$. Then $\frac{N h v}{10^{-6}}=10^{-10}$
or $N=\frac{10^{-6} \times 10^{-10}}{h v}=\frac{10^{-16} \lambda}{h c} $
$(\because c=v \lambda)$
$=\frac{10^{-16} \times\left(5.6 \times 10^{-7}\right)}{\left(6.6 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}$
$=0.28 \times 10^{3} \approx 3 \times 10^{2}$ photons / s