Question

A $60 \, W$ bulb is placed at a distance of $4 \, m$ from you. The bulb is emitting light of wavelength $600 \, nm$ uniformly in all directions. In $0.1 \, s$ , how many photons will enter your eye, if the pupil of the eye has a diameter of $2 \, mm$ ? [Take $hc=1240 \, eV \, nm$ ]

• A. 2.84 x 10¹²
• B. 2.84 x 10¹¹
• C. 9.37 x 10¹¹
• D. 6.84 x 10¹¹

Answer 1

Answer: (B)

The intensity of light at the location of your eye is
$\text{I} = \frac{\text{P}}{4 \pi \text{r}^{2}} = \frac{6 0}{4 \pi \times 4^{2}} W \text{m}^{- 2}$
The energy entering into your eye per second is
$\text{P}_{1} = \text{I} \times \frac{\pi \text{d}^{2}}{4}$
where d is the diameter of pupil
$\left(\text{P}\right)_{1} = \frac{6 0}{4 \pi \times 4^{2}} \times \frac{\pi \times \left(2 \times 1 0^{- 3}\right)^{2}}{4}$
$= \text{9.375} \times 1 0^{- 7} Js^{- 1}$
Let n be the number of photons entering into the eye per second, then
$\text{P}_{1} = \text{n} \times \frac{\text{hc}}{\lambda }$
$\text{9.375} \times 1 0^{- 7} = \text{n} \times \frac{1 2 4 0 \times \text{1.6} \times 1 0^{- 1 9}}{6 0 0}$
$\text{n} = \text{2.84} \times 1 0^{1 2} photon \text{s}^{- 1}$
So, the number of photons entering the eye in 0.1 s = 0.1 n = 2.84 x 10¹¹