A point source is placed at origin. Its intensity at distance of 2 cm from source is I then intensity at distance 4 cm from the source shall be.
\(\frac{I}{2}\)
\(\frac{I}{16}\)
\(\frac{I}{4}\)
- I
The Correct Option is C
Solution and Explanation
The problem asks for the intensity of light from a point source at a distance of 4 cm, given that its intensity is I at a distance of 2 cm.
Concept Used:
The intensity of light from a point source follows the inverse square law. This law states that the intensity (E) of the light is inversely proportional to the square of the distance (d) from the source. This is because a point source radiates energy uniformly in all directions, and as the distance increases, this energy spreads over the surface of an increasingly larger sphere.
The surface area of a sphere with radius \(d\) is \(A = 4\pi d^2\). If the source has a constant power P, the intensity (Power per unit area) is:
\[ E = \frac{P}{A} = \frac{P}{4\pi d^2} \]From this equation, we can see the relationship:
\[ E \propto \frac{1}{d^2} \]Step-by-Step Solution:
Step 1: Formulate the relationship between intensities at two different distances.
Let \(E_1\) be the intensity at a distance \(d_1\) and \(E_2\) be the intensity at a distance \(d_2\). From the inverse square law, we can write the ratio:
\[ \frac{E_2}{E_1} = \frac{1/d_2^2}{1/d_1^2} = \frac{d_1^2}{d_2^2} = \left(\frac{d_1}{d_2}\right)^2 \]Step 2: Identify the given values from the problem statement.
The initial intensity is \(E_1 = I\).
The initial distance is \(d_1 = 2 \, \text{cm}\).
The final distance is \(d_2 = 4 \, \text{cm}\).
We need to find the final intensity, \(E_2\).
Step 3: Substitute the known values into the ratio equation.
\[ \frac{E_2}{I} = \left(\frac{2 \, \text{cm}}{4 \, \text{cm}}\right)^2 \]Step 4: Calculate the final intensity \(E_2\).
First, simplify the fraction inside the parentheses:
\[ \frac{E_2}{I} = \left(\frac{1}{2}\right)^2 \]Now, square the fraction:
\[ \frac{E_2}{I} = \frac{1}{4} \]Finally, solve for \(E_2\):
\[ E_2 = \frac{I}{4} \]The intensity at a distance of 4 cm from the source shall be I/4.
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Concepts Used:
Sound
Sound is a vibration that propagates the mechanical wave of displacement and pressure, through a medium can be of any matter. In other words, the sound is the thin line between Music sound and Noise.
Types of Sounds:
Sound can be divided into two types depending on its frequency. The following are:-
Infrasonic Waves:
- Infrasonic waves have a frequency lower than 20 Hz. Since it is below the human hearing range, it is inaudible to humans.
- However, several other living organisms use these waves to communicate, whales, sharks, elephants, hippopotamus, etc. these waves are also used in the scientific field, it is used to study upcoming earthquakes, volcanic eruptions, human heart rates, rock formation stages, etc.
Ultrasonic waves:
- Ultrasonic waves have a frequency higher than 20 kHz. Since it is above the human hearing range, it is inaudible to humans.
- These waves are used in the scientific field, it is used to study human heart rates, to take sonograms, navigation of directions, imaging things, uniform sample mixing, communication, testing of samples, etc.