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Visible light of wavelength 6,000 × 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced at θ1, then  is close to
  • a)
    45°
  • b)
    30°
  • c)
    25°
  • d)
    20°
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Visible light of wavelength 6,000 × 10-8 cm falls normally on a ...
Given Data:
- Wavelength of light, λ = 6,000 × 10-8 cm
- Second diffraction minimum at 60° from central maximum

Calculating Angle of First Minimum:
- For a single slit, the condition for the first minimum is given by sinθ₁ = λ/a, where 'a' is the width of the slit.
- In this case, the angle of the second minimum is 60°, which means the first minimum is at half this angle (30°).

Calculating Angle of First Minimum:
- Substituting the values, sinθ₁ = λ/a = 6,000 × 10-8 / a
- As sin 30° = 1/2, we have 1/2 = 6,000 × 10-8 / a
- Solving for 'a', we get a = 12,000 × 10-8 cm = 1.2 × 10-4 cm

Final Calculation:
- Now, we need to calculate the angle 'α' that the first minimum makes with the central maximum.
- Using the small angle approximation, tanα = θ₁, where α is the angle with the central maximum.
- Therefore, tanα = tan30° ≈ 1/√3
- Hence, the angle 'α' is approximately 30°.
Therefore, the correct answer is option 'b) 30°'.
Free Test
Community Answer
Visible light of wavelength 6,000 × 10-8 cm falls normally on a ...
d sin θ = Path difference = 2λ
So,
d sin 60 = 2λ

For first minima,
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Visible light of wavelength 6,000 × 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced atθ1, then is close toa)45°b)30°c)25°d)20°Correct answer is option 'C'. Can you explain this answer?
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Visible light of wavelength 6,000 × 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced atθ1, then is close toa)45°b)30°c)25°d)20°Correct 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 Visible light of wavelength 6,000 × 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced atθ1, then is close toa)45°b)30°c)25°d)20°Correct 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 Visible light of wavelength 6,000 × 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60° from the central maximum. If the first minimum is produced atθ1, then is close toa)45°b)30°c)25°d)20°Correct answer is option 'C'. Can you explain this answer?.
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