Wave Optics

To solve the problem, we need to identify the correct characteristic of LEDs (Light Emitting Diodes) from the given statements.

1. Key Characteristics of LEDs:
- LEDs operate at low voltages and currents, making them energy-efficient.
- They do not require warm-up time; they light up instantly when current flows.
- The emission wavelength of LEDs depends on the semiconductor material used and is typically within a narrow range, not the full 4000–7000 Å (which is the visible light range in general).
- LEDs have very fast response times, allowing for rapid on-off switching.

2. Evaluating the Options:
- "High operational voltage" → Incorrect: LEDs work on low voltage.
- "Warm-up time is required" → Incorrect: LEDs are instant-on devices.
- "Bandwidth of light is 4000 Å – 7000 Å" → Incorrect: Each LED emits in a much narrower wavelength range.
- "Fast on-off switching" → Correct: This is a well-known and highly advantageous property of LEDs, used in digital displays and high-speed communication.

Final Answer:
The true statement about LEDs is: Fast on-off switching.

To solve the problem, we need to identify the correct statement about the propagation of waves.

1. Understanding Wave Types:
There are two main types of mechanical waves:
- Transverse Waves: In these waves, particle displacement is perpendicular to the direction of wave propagation. These waves require a medium with rigidity to propagate and generally do not travel through gases or liquids. They can propagate through solids.
- Longitudinal Waves: In these waves, particle displacement is parallel to the wave direction. These waves can propagate through solids, liquids, and gases.

2. Evaluate Each Option:
- Option (1): "Transverse waves cannot propagate through all media" – This is true, but not the best or most complete statement.
- Option (2): "Longitudinal waves can propagate through solids only" – This is incorrect, they can propagate through all states of matter.
- Option (3): "Transverse waves can propagate through solids" – This is correct and accurate.
- Option (4): "Longitudinal waves can propagate through vacuum" – This is false; mechanical waves (including longitudinal) cannot travel through a vacuum.

Final Answer:
Transverse waves can propagate through solids is the correct statement.

To solve the problem, we need to find the ratio , where is the root of the mean square of the x-component velocity of gas A molecules, and is the root mean square velocity of gas B molecules.

1. Understanding the Relation for RMS Velocity:
The root mean square (RMS) speed of gas molecules is given by:

Where:
- is Boltzmann constant
- is temperature
- is the mass of one molecule

2. Details from the Question:
- Gas A: N molecules, each of mass =
- Gas B: 2N molecules, each of mass =
- Temperature = (same for both gases)
- Mean square velocity of gas B =
- Mean square of x-component of velocity of gas A =

3. Mean Square of One Component of Velocity:
Since the velocity components are isotropic (equal in all directions), we know:

So,

4. Take Ratio of Speeds:
We want
Substitute the values:

So,

Final Answer:

To determine the ratio in Young's double slit experiment, we need to analyze the interference pattern and how the path difference affects the intensity of light at different points on the screen. 1. Given Data: - Wavelength of light, - Path difference at point 1, - Path difference at point 2, - Intensity at point 1, - Intensity at point 2, 2. Calculate the Phase Difference: - The phase difference is related to the path difference by: - For point 1: - For point 2: 3. Calculate the Intensity: - The intensity at a point on the screen is given by: where is the intensity of light from a single slit. - For point 1: - For point 2: 4. Determine the Ratio : - The ratio of the intensities is: 5. Final Answer: - The ratio is: This corresponds to option (1).

Step 1: Relation Between Frequency and Elastic Properties The fundamental frequency of the transverse wave in a stretched string is given by: where: - (length of the wire), - is the mass per unit length, given by: where is the density.

Step 2: Compute Tension Using Young's Modulus Given that strain is , and :

Step 3: Compute Frequency Substituting in the frequency formula: After solving, Thus, the correct answer is:

Step 1: Understanding the Reflection and Refraction of Light When a monochromatic light wave travels from one medium to another, it undergoes reflection and refraction at the interface.

Step 2: Property of Frequency in Different Media The frequency of light remains unchanged during reflection and refraction. This is because the frequency of a wave is determined by the source and is not affected by the medium through which it propagates.

Step 3: Effects on Other Parameters - The wavelength of light changes as it enters a new medium, because the velocity of light varies according to the refractive index of the medium. - The velocity of light also changes due to the medium's optical density. - The amplitude of light may change due to partial reflection and absorption at the interface. Thus, the correct answer is:

Step 1: Frequency of Fundamental Mode for a Closed Pipe For a closed pipe, the fundamental frequency is given by: where: - m/s (speed of sound in air), - is the length of the pipe.

Step 2: Compute Frequency for the Shorter Pipe For the shorter pipe of length 150 cm ( m):

Step 3: Determine Frequency of Longer Pipe Given the beat frequency is 6 Hz, the frequency of the longer pipe is:

Step 4: Compute Length of the Longer Pipe Using the formula for the fundamental frequency: Thus, the correct answer is: