Problem 42 - Master Resource Book in JEE Main Physics

Question

Two monochromatic beams \(A\) and \(B\) of equal intensity \(I\), hit a screen. The number of photons hitting the screen by beam \(A\) is twice that by beam \(B\). Then, what inference can you move about their frequencies? (a) The frequency of beam \(B\) is twice that of \(A\) (b) The frequency of beam \(B\) is half that of \(A\) (c) The frequency of beam \(A\) is twice of \(B\) (d) None of the above

Step-by-Step Solution

Answer

The frequency of beam \(B\) is twice that of \(A\). (Option (a))

Step 1: Understand the Relationship Between Intensity and Photon Number

The intensity of a monochromatic beam is related to the energy of the photons and the number of photons. For a beam of light, the total energy is given by \(E = n \cdot h \cdot f\), where \(n\) is the number of photons, \(h\) is Planck's constant, and \(f\) is the frequency.

Step 2: Express the Intensity in Terms of Photon Number and Frequency

Since the intensity \(I\) of both beams is equal and given by \(I = n \cdot h \cdot f / A\), where \(A\) is the area, and considering that the same area is covered for both beams, it can be expressed as \(n_A \cdot f_A = n_B \cdot f_B\).

Step 3: Apply the Given Condition

We are told that the number of photons hitting the screen from beam \(A\) is twice that of beam \(B\), or \(n_A = 2n_B\).

Step 4: Set Up the Equation for Frequency

From the relation in Step 2 and substituting \(n_A = 2n_B\), we can write \(2n_B \cdot f_A = n_B \cdot f_B\).

Step 5: Solve the Equation for the Frequency Ratio

Cancel out \(n_B\) to simplify: \(2f_A = f_B\). This implies \(f_A = \frac{f_B}{2}\).

Step 6: Make the Final Conclusion

Since \(f_A\) is half of \(f_B\), or equivalently \(f_B\) is twice of \(f_A\), the answer to the frequency relationship is (a): "The frequency of beam \(B\) is twice that of \(A\)."

Key Concepts

Intensity of Light

Light intensity is an important concept when studying electromagnetic waves. It tells us how much power is being distributed over a certain area. The more intense a light beam is, the more energy it carries.

For light, intensity is directly proportional to two things:

• Number of photons
• Energy of each photon

Energy per photon depends on the light's frequency. Therefore, if two beams of light have the same intensity but differ in the number of photons, one must have different frequencies. The energy equation for photons is:- Total Energy = Number of Photons \( \times \) Planck's constant \( \times \) Frequency
So, intensity can be thought of as a balance between how many photons hit an area and how much energy each one has. If a beam's photon number is high but it has low frequency, another beam can achieve the same intensity with fewer photons but higher frequency.

Frequency Relation

The frequency of light is a property that affects both its color and energy. In our exercise, we need to understand how frequency relates to the number of photons in beams with the same intensity. Given the equation for energy:- Energy = \( n \times h \times f \)we can see how frequency \( f \) plays a key role. Here are some important points:

• High frequency means each photon is energetic.
• Low frequency means each photon carries less energy.

For beams of equal intensity, if one beam has more photons, it must have lower frequency. If fewer photons are observed, they must be more energetic, thus with higher frequency. This is how frequency is related to the number of photons. The exercise shows that beam A has twice the photons than beam B, meaning it must have half the frequency of beam B.

Planck's Constant

Planck's constant is a fundamental part of physics that relates the energy of a photon to its frequency. It is a tiny value (\( h = 6.626 imes 10^{-34} ext{ Js} \)) and is crucial for understanding quantum mechanics. Here's why it's essential:- Planck's constant helps determine the energy carried by a single photon through the equation: \( E = h \times f \).This constant makes it possible to connect classical and quantum physics by relating energy and frequency. It tells us that higher frequency photons carry more energy. In our scenario from the exercise, with equal intensity but differing photon numbers, Planck's constant aids us in understanding the energy balancing act, highlighting the inverse relation between photon count and frequency.