JEE Exam > JEE Questions > In a double-slit experiment, at a certain poi...
Start Learning for Free
In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is
- a)0.568
- b)0.760
- c)0.853
- d)0.672
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
In a double-slit experiment, at a certain point on the screen, the pat...
Given:
- In a double-slit experiment
- Path difference between the two interfering waves at a certain point on the screen is 1/8th of a wavelength
To find:
- The ratio of the intensity of light at that point to that at the center of a bright fringe
Explanation:
The intensity of light in a double-slit experiment can be calculated using the formula:
I = 4 * I0 * cos^2(π * d * sinθ / λ)
Where:
- I is the intensity of light at a certain point on the screen
- I0 is the intensity of light at the center of a bright fringe
- d is the distance between the two slits
- θ is the angle between the line joining the point on the screen and the center of the pattern and the line joining the two slits
- λ is the wavelength of light
Step 1:
Calculate the path difference between the two interfering waves at the given point on the screen.
Path difference = 1/8th of a wavelength = λ/8
Step 2:
Using the path difference, calculate the phase difference between the two interfering waves.
Phase difference = 2π * path difference / wavelength = 2π * (λ/8) / λ = π/4
Step 3:
Calculate the angle θ using the formula:
sinθ = phase difference / distance between the point on the screen and the center of the pattern
Since sinθ is small for small angles, we can approximate sinθ ≈ θ in radians.
θ = phase difference / distance between the point on the screen and the center of the pattern
Step 4:
Substitute the calculated value of θ in the intensity formula to find the ratio of the intensity at the given point to the intensity at the center of a bright fringe.
I / I0 = 4 * cos^2(π * d * sinθ / λ) / 4 * cos^2(0) = cos^2(π * d * θ / λ) / cos^2(0) = cos^2(π * d * θ / λ)
Step 5:
Calculate the value of cos^2(π * d * θ / λ) using the given values and substitute in the formula.
I / I0 = cos^2(π * d * θ / λ) = cos^2(π * d * (π/4) / λ) = cos^2(π^2 * d / (4 * λ))
Step 6:
Since the value of d/λ is not given, we can assume d = λ/2 to simplify the calculation.
I / I0 = cos^2(π^2 * (λ/2) / (4 * λ)) = cos^2(π/8)
Step 7:
Using the trigonometric identity cos^2(x) = (1 + cos(2x)) / 2, we can simplify further.
I / I0 = (1 + cos(2 * (π/8))) / 2 = (1 + cos(π/4)) / 2
Step 8:
Using
- In a double-slit experiment
- Path difference between the two interfering waves at a certain point on the screen is 1/8th of a wavelength
To find:
- The ratio of the intensity of light at that point to that at the center of a bright fringe
Explanation:
The intensity of light in a double-slit experiment can be calculated using the formula:
I = 4 * I0 * cos^2(π * d * sinθ / λ)
Where:
- I is the intensity of light at a certain point on the screen
- I0 is the intensity of light at the center of a bright fringe
- d is the distance between the two slits
- θ is the angle between the line joining the point on the screen and the center of the pattern and the line joining the two slits
- λ is the wavelength of light
Step 1:
Calculate the path difference between the two interfering waves at the given point on the screen.
Path difference = 1/8th of a wavelength = λ/8
Step 2:
Using the path difference, calculate the phase difference between the two interfering waves.
Phase difference = 2π * path difference / wavelength = 2π * (λ/8) / λ = π/4
Step 3:
Calculate the angle θ using the formula:
sinθ = phase difference / distance between the point on the screen and the center of the pattern
Since sinθ is small for small angles, we can approximate sinθ ≈ θ in radians.
θ = phase difference / distance between the point on the screen and the center of the pattern
Step 4:
Substitute the calculated value of θ in the intensity formula to find the ratio of the intensity at the given point to the intensity at the center of a bright fringe.
I / I0 = 4 * cos^2(π * d * sinθ / λ) / 4 * cos^2(0) = cos^2(π * d * θ / λ) / cos^2(0) = cos^2(π * d * θ / λ)
Step 5:
Calculate the value of cos^2(π * d * θ / λ) using the given values and substitute in the formula.
I / I0 = cos^2(π * d * θ / λ) = cos^2(π * d * (π/4) / λ) = cos^2(π^2 * d / (4 * λ))
Step 6:
Since the value of d/λ is not given, we can assume d = λ/2 to simplify the calculation.
I / I0 = cos^2(π^2 * (λ/2) / (4 * λ)) = cos^2(π/8)
Step 7:
Using the trigonometric identity cos^2(x) = (1 + cos(2x)) / 2, we can simplify further.
I / I0 = (1 + cos(2 * (π/8))) / 2 = (1 + cos(π/4)) / 2
Step 8:
Using
Free Test
FREE
| Start Free Test |
Community Answer
In a double-slit experiment, at a certain point on the screen, the pat...
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer?
Question Description
In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer?.
In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer?.
Solutions for In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer?, a detailed solution for In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is 1/8th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe isa)0.568b)0.760c)0.853d)0.672Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup