A laser produces light of wavelength lambda. The light is directed through two splits that are separated by a small distance. On the other side of the slits is an interference pattern of alternating dark(black) and bright bands. What is the wavelength of the light in the bright bands?
2lambda
Lambda
0.5 lambda
9 lambda
A train passes a farm house at 30.0 m/s and then the engineer sounds the 900-Hz whistle. The air is still and the speed of sound is 340 m/s. What frequency will a person inside the farmhouse hear? Assume that the angle between the farmhouse and the train is negligible.
883 Hz
827 Hz
877 hz(my choice)
871 Hz
You are 15.0 m from the source of a sound. At that distance, you hear it at a sound level of 20.0 dB. How close must you move to the sound to increase the sound level to 60 dB?
15.0 cm
45.0 cm
55.0 cm
60.0 cm
85.0 cm
Please help. I’ll really appreciate it.
The interference pattern does not change the wavelength of the light ! Either this is a typo or the answer is lambda.
google doppler shift
340/370 * 900 = 827
logs are base 10
change in dB = 10 log P2/P1
40 = 10 log P2/P1
log P2/P1 = 4
P2/P1 = 10^4
the power goes square of distance inverted (same power through 4pi r^2)
distance ratio = 10^2 = 100
.15 meter = 15 cm
To calculate the wavelength of the light in the bright bands of an interference pattern created by a laser passing through two splits, you can use the concept of constructive interference. Constructive interference occurs when the waves from two different sources meet and their crests align, resulting in a brighter spot.
In this case, the bright bands in the interference pattern occur when the path length difference between the two splits is an integer multiple of the wavelength. Since the splits are separated by a small distance, we can assume that the path length difference is negligible compared to the wavelength.
Therefore, the wavelength of the light in the bright bands is equal to the distance between the two splits. Based on the given options, the wavelength of the light in the bright bands is Lambda.
Moving on to the second question about the frequency heard by a person inside a farmhouse as a train passes by at a speed of 30.0 m/s and the engineer sounds a 900-Hz whistle. The speed of sound is given as 340 m/s.
To calculate the frequency heard, we need to consider the Doppler effect. The Doppler effect causes a shift in frequency when there is relative motion between the source of the sound and the observer.
In this case, since the observer is inside the farmhouse and the angle between the farmhouse and the train is negligible, we can assume the relative motion between the source (train) and the observer (person inside the farmhouse) is only due to the train's velocity.
The formula to calculate the Doppler effect in this situation is given by:
Frequency heard = (Speed of sound + Velocity of observer) / (Speed of sound + Velocity of source) * Frequency of source
Plugging in the given values, we have:
Frequency heard = (340 + 0) / (340 + 30) * 900 = 827 Hz
Therefore, the frequency that a person inside the farmhouse will hear is 827 Hz.
Lastly, in order to determine how close you must move to a sound source to increase the sound level from 20.0 dB to 60 dB, you need to understand the concept of the inverse square law.
According to the inverse square law, the sound level decreases by 6 dB every time the distance from the source is doubled. In this case, we want to increase the sound level by 40 dB (60 dB - 20 dB).
Since the sound level decreases by 6 dB when the distance doubles, to increase the sound level by 40 dB, we need to reduce the distance by a factor of:
10^(40/6) ≈ 12.59
Therefore, you would need to move to a distance approximately 1/12.59 = 0.0793 times the original distance (15.0 m) to increase the sound level to 60 dB.
Converting this distance to centimeters, we get:
0.0793 * 100 = 7.93 cm
Therefore, you would need to move approximately 7.93 cm closer to the sound source to increase the sound level to 60 dB.