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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
- a)0.32 mm
- b)0.2 mm
- c)0.3 mm
- d)0.25 mm
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A parallel beam of light of wavelength 500 nm falls on a narrow slit a...
Wavelength of light beam, λ
Distance of the screen from the slit, D=1m
For first minima, n=1
Distance between the slits is d
Distance of the first minimum from the centre of the screen can be obtained as, x = 2.5mm = 2.5×10−3
Now, nλ = xd/D
⇒ d= nλD/x = 0.2mm
Therefore, the width of the slits is 0.2 mm.
Most Upvoted Answer
A parallel beam of light of wavelength 500 nm falls on a narrow slit a...
Given:
Wavelength of light, λ = 500 nm = 500 × 10⁻⁹ m
Distance of screen from the slit, D = 1 m
Distance of first minimum from the centre of the screen, x = 2.5 mm = 2.5 × 10⁻³ m
To find: Width of the slit, a
Formula used:
The position of the nth minimum (on either side of the central bright fringe) is given by,
a sin θ = nλ
where,
a = width of the slit
n = order of diffraction minimum
λ = wavelength of light
θ = angle of diffraction (small angle approximation)
Calculation:
We know that for the first minimum, n = 1.
Using the small angle approximation, we can write sin θ ≈ θ, where θ is in radians.
Therefore, aθ = nλ
θ = x/D [Distance of first minimum from the centre of the screen = Distance of the screen from the slit × tan(θ)]
∴ a = nλθ [Substituting the value of θ]
a = (1)(500 × 10⁻⁹)(2.5 × 10⁻³)/(1)
a = 1.25 × 10⁻⁶ m = 0.125 mm
Therefore, the width of the slit is 0.2 mm (approx).
Hence, option (b) is the correct answer.
Wavelength of light, λ = 500 nm = 500 × 10⁻⁹ m
Distance of screen from the slit, D = 1 m
Distance of first minimum from the centre of the screen, x = 2.5 mm = 2.5 × 10⁻³ m
To find: Width of the slit, a
Formula used:
The position of the nth minimum (on either side of the central bright fringe) is given by,
a sin θ = nλ
where,
a = width of the slit
n = order of diffraction minimum
λ = wavelength of light
θ = angle of diffraction (small angle approximation)
Calculation:
We know that for the first minimum, n = 1.
Using the small angle approximation, we can write sin θ ≈ θ, where θ is in radians.
Therefore, aθ = nλ
θ = x/D [Distance of first minimum from the centre of the screen = Distance of the screen from the slit × tan(θ)]
∴ a = nλθ [Substituting the value of θ]
a = (1)(500 × 10⁻⁹)(2.5 × 10⁻³)/(1)
a = 1.25 × 10⁻⁶ m = 0.125 mm
Therefore, the width of the slit is 0.2 mm (approx).
Hence, option (b) is the correct answer.
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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.a)0.32 mmb)0.2 mmc)0.3 mmd)0.25 mmCorrect answer is option 'B'. Can you explain this answer?
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