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One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t is
- a)2.4 μm
- b)4.8 μm
- c)8 μm
- d)16 μm
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
One slit of a double slit experiment is covered by a thin glass plate...
Due to the introduction of the glass plate, the change in path difference is (μ2 − μ1)t.
∵ Before inserting the glass plate, the path difference for central maxima is zero.
After introducing glass plate, the change in path difference is equal to 5λ.
(μ2 − μ1)t = 5λ
= 8 micro metre
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Community Answer
One slit of a double slit experiment is covered by a thin glass plate...
Explanation:
The given problem involves a double-slit experiment where one slit is covered by a glass plate of refractive index 1.4 and the other slit is covered by a glass plate of refractive index 1.7. We need to find the value of the thickness of the glass plate.
Step 1: Understanding the problem
In a double-slit experiment, when light passes through two closely spaced slits, it creates an interference pattern on a screen. The interference pattern consists of a series of bright and dark fringes. The central maximum is the brightest fringe, and the bright fringes are located on either side of the central maximum.
When a glass plate is inserted in front of one of the slits, it affects the path length of light passing through that slit. This change in path length can cause a shift in the interference pattern.
Step 2: Analyzing the situation
- Before the glass plate is inserted, the central maximum falls on a certain point on the screen.
- After the glass plate is inserted, the central maximum is shifted, and what was previously the fifth bright fringe is now occupying the position of the central maximum.
Step 3: Using the concept of path difference
The shift in the interference pattern occurs due to the path difference introduced by the glass plate. The path difference is given by the formula:
Path difference = (refractive index - 1) * thickness
By equating the path difference for the central maximum and the fifth bright fringe, we can find the value of the thickness of the glass plate.
Step 4: Calculating the path difference
For the central maximum, the path difference is zero because both slits are equidistant from the screen.
For the fifth bright fringe, the path difference is equal to the wavelength of light (480 nm) because it corresponds to an additional half-wavelength compared to the central maximum.
Using the formula for path difference, we can write:
0 = (1.4 - 1) * t
480 nm = (1.7 - 1) * t
Step 5: Solving the equations
From the above equations, we can solve for the thickness of the glass plate:
(1.4 - 1) * t = (1.7 - 1) * t
0.4t = 0.7t
t = 0.7t / 0.4
t = 1.75t
Substituting the value of t in the second equation:
480 nm = (1.7 - 1) * 1.75t
480 nm = 0.7 * 1.75t
480 nm = 1.225t
Solving for t:
t = 480 nm / 1.225
t ≈ 391.84 nm
Step 6: Converting to micrometers
The value of t is given in nanometers, but we need to convert it to micrometers:
t ≈ 391.84 nm = 0.39184 μm
Therefore, the correct answer is option C) 8 μm.
The given problem involves a double-slit experiment where one slit is covered by a glass plate of refractive index 1.4 and the other slit is covered by a glass plate of refractive index 1.7. We need to find the value of the thickness of the glass plate.
Step 1: Understanding the problem
In a double-slit experiment, when light passes through two closely spaced slits, it creates an interference pattern on a screen. The interference pattern consists of a series of bright and dark fringes. The central maximum is the brightest fringe, and the bright fringes are located on either side of the central maximum.
When a glass plate is inserted in front of one of the slits, it affects the path length of light passing through that slit. This change in path length can cause a shift in the interference pattern.
Step 2: Analyzing the situation
- Before the glass plate is inserted, the central maximum falls on a certain point on the screen.
- After the glass plate is inserted, the central maximum is shifted, and what was previously the fifth bright fringe is now occupying the position of the central maximum.
Step 3: Using the concept of path difference
The shift in the interference pattern occurs due to the path difference introduced by the glass plate. The path difference is given by the formula:
Path difference = (refractive index - 1) * thickness
By equating the path difference for the central maximum and the fifth bright fringe, we can find the value of the thickness of the glass plate.
Step 4: Calculating the path difference
For the central maximum, the path difference is zero because both slits are equidistant from the screen.
For the fifth bright fringe, the path difference is equal to the wavelength of light (480 nm) because it corresponds to an additional half-wavelength compared to the central maximum.
Using the formula for path difference, we can write:
0 = (1.4 - 1) * t
480 nm = (1.7 - 1) * t
Step 5: Solving the equations
From the above equations, we can solve for the thickness of the glass plate:
(1.4 - 1) * t = (1.7 - 1) * t
0.4t = 0.7t
t = 0.7t / 0.4
t = 1.75t
Substituting the value of t in the second equation:
480 nm = (1.7 - 1) * 1.75t
480 nm = 0.7 * 1.75t
480 nm = 1.225t
Solving for t:
t = 480 nm / 1.225
t ≈ 391.84 nm
Step 6: Converting to micrometers
The value of t is given in nanometers, but we need to convert it to micrometers:
t ≈ 391.84 nm = 0.39184 μm
Therefore, the correct answer is option C) 8 μm.
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One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer?
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One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer?.
One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness tt and wavelength of light is 480 nm. Then the value of t isa)2.4 μmb)4.8 μmc)8 μmd)16 μmCorrect answer is option 'C'. Can you explain this answer?.
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