A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Powe

Q: Wait, why does light have momentum if it has no mass?

Great question! In classical mechanics, $p = mv$, so zero mass means zero momentum. But in Einstein's relativity, the full equation is $E^2 = (pc)^2 + (mc^2)^2$. For a photon, the rest mass $m$ is zero, which simplifies the equation to $E = pc$, or $p = \frac{E}{c}$.

Q: Why do we assume all the momentum is transferred to the object?

The problem explicitly states that the pulse is "absorbed completely". This is perfectly analogous to a completely inelastic collision in mechanics. The object "catches" the photons and keeps all their momentum.

Q: What if the light was reflected instead of absorbed?

If the object was a perfect mirror and reflected the light straight back, the change in the light's momentum would be $p_{final} - p_{initial} = -p - p = -2p$. By conservation of momentum, the object would gain $2p$, exactly double the momentum!

Q: Why can't I just multiply 30 and 100 directly?

This is a classic JEE trap! The values are given in milliwatts (mW) and nanoseconds (ns). To get the final answer in standard SI units ($\text{kg-m/s}$), you must convert everything to base SI units: Watts and seconds.

Q: Is this tiny amount of momentum actually useful for anything?

Yes! While $10^{-17} \text{ kg-m/s}$ is imperceptible to us, in the vacuum of space, continuous exposure to sunlight can provide a steady acceleration. This is the exact principle behind "solar sails" used in experimental spacecraft!

Question

A pulse of light of duration $100 	ext{ ns}$ is absorbed completely by a small object initially at rest. Power of the pulse is $30 	ext{ mW}$ and the speed of light is $3 	imes 10^8 	ext{ ms}^{-1}$. The final momentum of the object is

Options:

(A) $0.3 	imes 10^{-17} 	ext{ kg-ms}^{-1}$

(B) $1.0 	imes 10^{-17} 	ext{ kg-ms}^{-1}$

(C) $3.0 	imes 10^{-17} 	ext{ kg-ms}^{-1}$

(D) $9.0 	imes 10^{-17} 	ext{ kg-ms}^{-1}$

Sigma Percentile · Accepted Answer

### The Invisible Push

Have you ever stood in the sunlight and felt it pushing you? Probably not. But light, despite having no mass, carries momentum. When a photon strikes an object and is absorbed, it transfers this momentum, giving the object a microscopic push. In this problem, we are going to calculate exactly how much momentum a tiny pulse of light imparts to an object.

### The Master Equation

To solve this, we need to bridge the gap between classical mechanics and modern physics. For a particle with mass, momentum is simply mass times velocity. But for light, we use the relativistic relationship:

$$p = \frac{E}{c}$$

where $p$ is the momentum, $E$ is the total energy of the light pulse, and $c$ is the speed of light.

However, the problem doesn't hand us the energy on a silver platter. Instead, it gives us the **power** of the pulse and its **duration**. We know that power is the rate at which energy is delivered. Therefore, the total energy can be found by multiplying power by time:

$$E = P 	imes t$$

### Combining the Concepts

By substituting our energy expression into the momentum formula, we get a direct equation for the momentum transferred:

$$p = \frac{P 	imes t}{c}$$

This is the exact arsenal we need. Now, it's just a matter of plugging in the numbers. But beware—this is where many students fall into the trap of ignoring units!

### The Execution

We are given:
- Power, $P = 30 	ext{ mW} = 30 	imes 10^{-3} 	ext{ W}$
- Time, $t = 100 	ext{ ns} = 100 	imes 10^{-9} 	ext{ s}$
- Speed of light, $c = 3 	imes 10^8 	ext{ m/s}$

Let's first calculate the total energy $E$:

$$E = (30 	imes 10^{-3}) 	imes (100 	imes 10^{-9})$$

$$E = 3000 	imes 10^{-12} = 3 	imes 10^{-9} 	ext{ J}$$

Now, we divide this energy by the speed of light to find the momentum:

$$p = \frac{3 	imes 10^{-9}}{3 	imes 10^8}$$

Notice how beautifully the $3$ in the numerator and denominator cancel out. We are left with:

$$p = 1.0 	imes 10^{-17} 	ext{ kg-ms}^{-1}$$

This incredibly small number represents the momentum imparted to the object. While it might seem insignificant, this exact principle is what engineers use to design solar sails, allowing spacecraft to be propelled through the cosmos using nothing but the push of sunlight!

Animated Solution for Physics - Dual Nature of Matter and Radiation: A pulse of light of duration $100 ns$ is absorbed completely by a small object initially at rest. Power of the pulse is $30 mW$ and the speed of light is $3 \times 1 0^{8} ms^{- 1}$ . The final momentum of the object is

Select Answer:

Visualized Solution

The Physical Setup

$p = \frac{E}{c}$

$E = P \times t$

$p = \frac{P \times t}{c}$

Substituting Values

Calculating Energy

Calculating Momentum

Final Answer

The Sigma Insight: Photon Theory of Light

The Invisible Push

The Master Equation

Combining the Concepts

The Execution

Similar Questions

Photon of frequency $ν$ has a momentum associated with it. If $c$ is the velocity of light, the momentum is

A $2 mW$ laser operates at a wavelength of $500 nm$ . The number of photons that will be emitted per second is [Given, Planck's constant $h = 6.6 \times 1 0^{- 34} Js$ , speed of light $c = 3.0 \times 1 0^{8} m/s$ ]

The wavelength of an X-ray beam is $10 \overset{˚}{A}$ . The mass of a fictitious particle having the same energy as that of the X-ray photons is $\frac{x}{3} h kg$ . The value of $x$ is ............ . ( $h = Planck’s constant$ )

Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is

If a source of power $4 kW$ produces $1 0^{20}$ photons/second, the radiation belong to a part of the spectrum called

An X-ray tube is operated at $1.24$ million volt. The shortest wavelength of the produced photon will be

Four physical quantities are listed in Column I. Their values are listed in Column II in a random order

List-I

List-II

X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from