JEE Exam > JEE Questions > In a Youngs double slit experiment, the separ...
Start Learning for Free
In a Young's double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:
- a)3.9 mm
- b)6.9 mm
- c)5.9 mm
- d)4.9 mm
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 Youngs double slit experiment, the separation between the slits i...
To find the separation between the successive bright fringes on the screen in a Young's double slit experiment, we can use the formula:
δy = λL/d
Where:
δy is the separation between the bright fringes on the screen,
λ is the wavelength of light used,
L is the distance between the double slits and the screen,
and d is the separation between the double slits.
Given:
λ = 589 nm (or 589 × 10^-9 m),
L = 1.5 m,
and d = 0.15 mm (or 0.15 × 10^-3 m).
Let's substitute these values into the formula to find the separation between the bright fringes:
δy = (589 × 10^-9 m) × (1.5 m) / (0.15 × 10^-3 m)
= (589 × 1.5) / 0.15 × 10^-6
= 883.5 × 10^-6 / 0.15 × 10^-6
= 883.5 / 0.15
≈ 5890 mm
Therefore, the separation between the successive bright fringes on the screen is approximately 5890 mm.
Now let's convert this to the given options:
a) 3.9 mm
b) 6.9 mm
c) 5.9 mm
d) 4.9 mm
We can see that the correct answer is option c) 5.9 mm.
Therefore, the separation between the successive bright fringes on the screen is 5.9 mm.
δy = λL/d
Where:
δy is the separation between the bright fringes on the screen,
λ is the wavelength of light used,
L is the distance between the double slits and the screen,
and d is the separation between the double slits.
Given:
λ = 589 nm (or 589 × 10^-9 m),
L = 1.5 m,
and d = 0.15 mm (or 0.15 × 10^-3 m).
Let's substitute these values into the formula to find the separation between the bright fringes:
δy = (589 × 10^-9 m) × (1.5 m) / (0.15 × 10^-3 m)
= (589 × 1.5) / 0.15 × 10^-6
= 883.5 × 10^-6 / 0.15 × 10^-6
= 883.5 / 0.15
≈ 5890 mm
Therefore, the separation between the successive bright fringes on the screen is approximately 5890 mm.
Now let's convert this to the given options:
a) 3.9 mm
b) 6.9 mm
c) 5.9 mm
d) 4.9 mm
We can see that the correct answer is option c) 5.9 mm.
Therefore, the separation between the successive bright fringes on the screen is 5.9 mm.
Free Test
FREE
| Start Free Test |
Community Answer
In a Youngs double slit experiment, the separation between the slits i...
Fringe-width,
= 589 × 10-2 mm = 5.89 mm
≈ 5.9 mm
≈ 5.9 mm
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 Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer?
Question Description
In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect 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 Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect 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 Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer?.
In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect 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 Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect 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 Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer?.
Solutions for In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect 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 Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a Youngs double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern observed on a screen kept is 1.5 m away. The separation between the successive bright fringes on the screen is:a)3.9 mmb)6.9 mmc)5.9 mmd)4.9 mmCorrect 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