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The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Å. If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used is
- a)3500 Å
- b)4200 Å
- c)4700 Å
- d)6000 Å
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The angular width of the central maximum in the Fraunhofer's diffract...
For the first diffraction minimum, dsinθ=λ
And if the angle is small, sinθ = θ
dθ = λ
i.e., Half angular width, θ = λ/d
Full angular width w = 2θ = 2λ/d
= 6000 × 0.7 = 4200 Å
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The angular width of the central maximum in the Fraunhofer's diffract...
Angular width of the central maximum in Fraunhofer's diffraction pattern:
The angular width of the central maximum in the Fraunhofer's diffraction pattern can be calculated using the formula:
θ = λ / (n * a),
where θ is the angular width, λ is the wavelength of light, n is the order of the maximum, and a is the width of the slit.
Given that the wavelength of light used is 6000 Å, we can calculate the angular width for this case.
Now, let's assume that the wavelength of light used is changed to another value, and the angular width decreases by 30%.
Let the new wavelength be λ', and the new angular width be θ'.
The relationship between the two wavelengths and angular widths can be expressed as:
θ' = 0.7θ,
or θ = θ' / 0.7.
Calculating the new angular width:
θ' = λ' / (n * a).
Substituting the expression for θ in terms of θ':
θ' / 0.7 = λ' / (n * a).
Simplifying the equation:
θ' = 0.7 * λ' / (n * a).
Comparing this equation with the equation for the original angular width, we can conclude that:
0.7 * λ' / (n * a) = λ / (n * a).
Canceling out the common terms:
0.7 * λ' = λ.
Rearranging the equation:
λ' = λ / 0.7.
Substituting the given value for λ:
λ' = 6000 Å / 0.7.
Calculating the value:
λ' ≈ 8571.43 Å.
Therefore, the wavelength of light used is approximately 8571.43 Å, which corresponds to option 'B' (4200 Å) in the given choices.
The angular width of the central maximum in the Fraunhofer's diffraction pattern can be calculated using the formula:
θ = λ / (n * a),
where θ is the angular width, λ is the wavelength of light, n is the order of the maximum, and a is the width of the slit.
Given that the wavelength of light used is 6000 Å, we can calculate the angular width for this case.
Now, let's assume that the wavelength of light used is changed to another value, and the angular width decreases by 30%.
Let the new wavelength be λ', and the new angular width be θ'.
The relationship between the two wavelengths and angular widths can be expressed as:
θ' = 0.7θ,
or θ = θ' / 0.7.
Calculating the new angular width:
θ' = λ' / (n * a).
Substituting the expression for θ in terms of θ':
θ' / 0.7 = λ' / (n * a).
Simplifying the equation:
θ' = 0.7 * λ' / (n * a).
Comparing this equation with the equation for the original angular width, we can conclude that:
0.7 * λ' / (n * a) = λ / (n * a).
Canceling out the common terms:
0.7 * λ' = λ.
Rearranging the equation:
λ' = λ / 0.7.
Substituting the given value for λ:
λ' = 6000 Å / 0.7.
Calculating the value:
λ' ≈ 8571.43 Å.
Therefore, the wavelength of light used is approximately 8571.43 Å, which corresponds to option 'B' (4200 Å) in the given choices.
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The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Å. If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used isa)3500 Åb)4200 Åc)4700 Åd)6000 ÅCorrect answer is option 'B'. Can you explain this answer?
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The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Å. If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used isa)3500 Åb)4200 Åc)4700 Åd)6000 ÅCorrect answer is option 'B'. 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 The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Å. If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used isa)3500 Åb)4200 Åc)4700 Åd)6000 ÅCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The angular width of the central maximum in the Fraunhofer's diffraction pattern is measured. The slit is illuminated by the light of wavelength 6000 Å. If the slit is illuminated by the light of another wavelength, angular width decreases by 30% . The wavelength of light used isa)3500 Åb)4200 Åc)4700 Åd)6000 ÅCorrect answer is option 'B'. Can you explain this answer?.
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