A laser light of wavelength 660 nm is used to weld Retina detachment. If a laser pulse of width 60 ms and power 0.5 kW is used, the approximate number of photons in the pulse are (Take Planck's Constant, $$h = 6.62 \times 10^{-34}$$ J s)

We are given the wavelength of the laser light as $$\lambda = 660 \text{ nm}$$. We first convert this nanometre value into metre because SI units must be used in every formula. Now, $$1 \text{ nm}=10^{-9}\text{ m}$$, so

$$\lambda = 660 \times 10^{-9}\text{ m}=6.60 \times 10^{-7}\text{ m}.$$

The laser power is given as $$0.5\text{ kW}$$. Knowing that $$1\text{ kW}=10^{3}\text{ W}$$, we rewrite

$$P = 0.5 \times 10^{3}\text{ W}=5.0 \times 10^{2}\text{ W}.$$

The pulse width (time for which the laser is ON) is $$60\text{ ms}$$. Converting milliseconds to seconds, $$1\text{ ms}=10^{-3}\text{ s}$$, so

$$\Delta t = 60 \times 10^{-3}\text{ s}=6.0 \times 10^{-2}\text{ s}.$$

First we calculate the total energy delivered in one pulse. The relation connecting power, energy and time is

$$E_{\text{total}} = P \, \Delta t.$$

Substituting the values, we have

$$E_{\text{total}} = \left(5.0 \times 10^{2}\text{ W}\right)\left(6.0 \times 10^{-2}\text{ s}\right) = 30\text{ J}.$$

Next we find the energy carried by a single photon of wavelength $$\lambda$$. The Planck-Einstein relation states

$$E_{\text{photon}} = \dfrac{h c}{\lambda},$$

where $$h = 6.62 \times 10^{-34}\text{ J s}$$ is Planck’s constant and $$c = 3.0 \times 10^{8}\text{ m s}^{-1}$$ is the speed of light in vacuum. Substituting, we get

$$E_{\text{photon}} = \dfrac{\left(6.62 \times 10^{-34}\text{ J s}\right)\left(3.0 \times 10^{8}\text{ m s}^{-1}\right)} {6.60 \times 10^{-7}\text{ m}} = \dfrac{19.86 \times 10^{-26}\text{ J m}} {6.60 \times 10^{-7}\text{ m}}.$$

Dividing the numbers and taking care of the powers of ten,

$$E_{\text{photon}} = 3.0 \times 10^{-19}\text{ J}.$$

The number of photons in one pulse is simply the ratio of the total pulse energy to the energy per photon:

$$N = \dfrac{E_{\text{total}}}{E_{\text{photon}}} = \dfrac{30\text{ J}}{3.0 \times 10^{-19}\text{ J}} = 10 \times 10^{19} = 1.0 \times 10^{20}.$$

This value is of the order of $$10^{20}$$ photons.

Hence, the correct answer is Option C.