Waves

Using the Doppler effect: Both source and observer are moving towards each other, so: Wavelength observed:

A progressive wave has the general form: Let’s analyze each: - (A) is a product of functions in and — represents a stationary wave, not progressive.
- (B) is not a harmonic wave (not sinusoidal).
- (C) Combine terms:
This represents a progressive wave.
- (D) Sum of two standing wave expressions — not a single progressive wave.

The time between two successive minima in a beat is given by the formula: Where and are the frequencies producing the beats. Thus, the correct expression is:

We are given the speed of the wave , the frequency , and the distance between the points . We can calculate the wavelength using the formula: Now, the phase difference between two points separated by a distance is given by: Substitute cm and cm: Thus, the phase difference is , so the correct answer is .

Step 1: Determine the wave number and wavelength.
Comparing with , we get , so .
Step 2: Identify the distance between a node and an adjacent antinode.
The distance between a node and an adjacent antinode in a stationary wave is .
Step 3: Calculate the distance.
Distance
Step 4: Convert to centimeters. Distance

Step 1: Identify the eye defect.
The normal least distance of distinct vision (near point) is . For the given boy, the least distance of distinct vision is . This means he cannot see objects clearly closer than . This defect is hypermetropia (farsightedness), where the eye lens converges light too weakly, causing the image to form behind the retina.
Step 2: Determine the type of lens required.
To correct hypermetropia, a convex lens is used. A convex lens converges the light rays before they enter the eye, allowing the eye's lens to focus the image correctly on the retina.
Step 3: Calculate the focal length of the corrective lens.
For a hypermetropic eye, the lens needs to form a virtual image of an object placed at the normal near point ( ) at the person's defective near point ( ). We use the lens formula:

Here, (object at normal near point)
(virtual image formed at the defective near point)
Substitute the values into the lens formula:


To add these fractions, find a common denominator, which is .



Therefore, the focal length is:


Since the focal length is positive, it confirms that a convex lens is required.

For a closed pipe (one end closed), the fundamental frequency is , where is the speed of sound and is the length.
For an open pipe (both ends open), the fundamental frequency is .
Given: Beat frequency = Hz.
So, .
When lengths are halved, new length .
New frequencies: Hz (since ), and Hz.
New beat frequency = Hz.

Einstein’s photoelectric equation: , where is the work function.
Case 1: Wavelength , (Equation 1).
Case 2: Wavelength , (Equation 2).
Subtract Equation 1 from Equation 2: .
Simplify: .
So, .
Substitute into Equation 1: .
Solve for : .

- For a pipe closed at one end and open at the other (closed pipe), the fundamental frequency corresponds to the first harmonic. The length of the pipe is related to the wavelength by:
- The speed of sound , frequency , and wavelength are related by:
- Given , :
- Thus, the length of the pipe:
- This matches option (A).

The number of beats per second is given by the difference in frequencies of the two tuning forks: where and . Now, when the frequencies of the sources are close but not exactly equal, destructive interference occurs at the beat minima, which happens when the sound waves are out of phase. Thus, the number of beats heard is 4 beats/s, and destructive interference occurs at the beat minima.
Final answer
Answer:

Let the frequency of the string be . Given: The common value from both sets is:

The general form of a travelling wave is: where is the amplitude, is the wave number (in rad/m), is the angular frequency (in rad/s), and is the phase constant. For the given wave: Compare with the general form: - Amplitude . - Wave number . - Angular frequency . The frequency (in Hz) is related to the angular frequency by: Solve for : Verify other parameters for context: - Period: . - Wavelength: . - Wave speed: . The frequency matches option (4). Check options: - (1) 6 Hz: , incorrect. - (2) 7 Hz: , incorrect. - (3) 4 Hz: , incorrect. - (4) 5 Hz: Correct. Thus, the frequency is 5 Hz, confirming option (4).

The phase difference between two waves determines their path difference. For the waves: The phase of is , and the phase of is . The phase difference is: The relationship between phase difference and path difference is: Solve for : Check options: - (1) : Incorrect, as path difference is proportional to , not its absolute value. - (2) : Incorrect, introduces an arbitrary phase shift. - (3) : Correct. - (4) : Incorrect, manipulates unnecessarily. Thus, the path difference is , confirming option (3).

Step 1: Understand the role of the ionosphere in radio communication.
The ionosphere is a part of Earth's upper atmosphere, approximately from 60 km to 1000 km above the surface, where solar and cosmic radiation ionize atoms and molecules, creating a plasma containing free electrons and ions. These ionized layers have the property of reflecting certain frequencies of electromagnetic waves (radio waves) back towards Earth. This phenomenon is crucial for long-distance radio communication, known as skywave propagation, as it allows signals to travel beyond the line of sight.
Step 2: Identify the typical frequency range for ionospheric reflection.
The ability of the ionosphere to reflect radio waves is frequency-dependent.
Very Low Frequencies (VLF) and Low Frequencies (LF) (e.g., kHz range): These waves are largely absorbed by the lower layers of the ionosphere (D-layer) during the day, though some reflection can occur at night. They are less efficiently reflected for typical long-distance communication compared to higher frequencies.
Medium Frequencies (MF) (e.g., hundreds of kHz to a few MHz): These waves also experience significant absorption, especially in the D-layer during the day, but can be reflected by the E and F layers, particularly at night.
High Frequencies (HF) (3 MHz to 30 MHz): This is the primary frequency range for which the ionosphere acts as a reliable reflector. These waves can penetrate the lower D and E layers and are efficiently reflected by the higher F-layer, enabling worldwide radio communication (shortwave radio). This is the "skywave" propagation that allows signals to bounce between the ionosphere and the Earth's surface over long distances.
Very High Frequencies (VHF) and Ultra High Frequencies (UHF) (above 30 MHz, into GHz range): These frequencies are generally too high to be reflected by the ionosphere. They typically pass through the ionosphere into space, making them suitable for satellite communication and line-of-sight terrestrial communication.
Step 3: Evaluate the given options.
(1) 3 - 30 kHz: This is in the VLF/LF range. While some interaction occurs, it's not the primary range for effective reflection for general communication.
(2) 3 - 30 MHz: This is the High Frequency (HF) band, which is precisely the range known for efficient ionospheric reflection and skywave propagation.
(3) 3 - 30 Hz: This is an extremely low frequency (ELF) range, used for very specialized applications (like communication with submarines) due to its unique propagation characteristics, not ionospheric reflection for typical radio.
(4) 3 - 30 GHz: This is in the microwave range (UHF and higher). Waves at these frequencies generally penetrate the ionosphere.
Step 4: Conclusion.
Based on the characteristics of radio wave propagation through the ionosphere, the frequency range of 3 - 30 MHz (High Frequency band) is where the ionosphere most effectively acts as a reflector for long-distance communication.
The final answer is .

Step 1: Recall the electromagnetic spectrum.
The electromagnetic (EM) spectrum is the range of all types of EM radiation. Radiation is energy that travels and spreads out as it goes – the visible light that comes from a lamp in your house and the radio waves that come from a radio station are two types of electromagnetic radiation. The only difference among them is their wavelength and frequency (and thus energy).
The EM spectrum, ordered by decreasing wavelength (and increasing frequency/energy), is as follows:
1. Radio waves
2. Microwaves
3. Infrared (IR)
4. Visible light
5. Ultraviolet (UV)
6. X-rays
7. Gamma rays
Step 2: Compare the wavelengths of the given options.
Let's look at the position of each given option in the spectrum:
X-rays: Located at the shorter wavelength end of the spectrum, with very high energy.
Radio waves: Located at the longest wavelength end of the spectrum, with very low energy.
UV waves (Ultraviolet): Located just beyond the violet end of the visible spectrum, having shorter wavelengths than visible light.
Visible rays (Visible light): Occupies a small portion in the middle of the spectrum, between infrared and ultraviolet.
Step 3: Identify the wave with the maximum wavelength.
Based on the order of the electromagnetic spectrum, Radio waves have the longest wavelength among all types of electromagnetic radiation. Therefore, among the given options (X-rays, Radio waves, UV waves, Visible rays), radio waves have the maximum wavelength. The final answer is .