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A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-
Correct answer is '0.15'. Can you explain this answer?
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A linear aperture whose width is 0.02 cm is placed immediately in fron...
f = D = 60 cm
For first minima,
For first minima,
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A linear aperture whose width is 0.02 cm is placed immediately in fron...
We need to use the formula for diffraction by a single slit:
d sin θ = mλ
where d is the width of the slit, θ is the angle of diffraction, m is the order of the diffraction maximum (m = 0 for the central maximum, m = ±1 for the first-order maxima, etc.), and λ is the wavelength of the light.
In this case, the width of the slit is d = 0.02 cm = 0.0002 m, the wavelength of the light is λ = 5 × 10^-7 m, and we want to find the position of the first-order maximum on the screen, which corresponds to m = ±1.
The angle of diffraction θ can be approximated by θ ≈ tan θ = y/L, where y is the distance from the center of the screen to the position of the maximum, and L is the distance from the slit to the screen. We can assume that the screen is far enough away that the distance from the lens to the screen is approximately equal to the focal length of the lens (i.e., L ≈ f = 60 cm).
For the first-order maximum, we have:
d sin θ = λ
0.0002 sin θ = 5 × 10^-7
sin θ ≈ 0.0025
θ ≈ 0.144 degrees
y/L ≈ 0.0025
y ≈ 0.15 cm
So the position of the first-order maximum is approximately 0.15 cm from the center of the screen.
d sin θ = mλ
where d is the width of the slit, θ is the angle of diffraction, m is the order of the diffraction maximum (m = 0 for the central maximum, m = ±1 for the first-order maxima, etc.), and λ is the wavelength of the light.
In this case, the width of the slit is d = 0.02 cm = 0.0002 m, the wavelength of the light is λ = 5 × 10^-7 m, and we want to find the position of the first-order maximum on the screen, which corresponds to m = ±1.
The angle of diffraction θ can be approximated by θ ≈ tan θ = y/L, where y is the distance from the center of the screen to the position of the maximum, and L is the distance from the slit to the screen. We can assume that the screen is far enough away that the distance from the lens to the screen is approximately equal to the focal length of the lens (i.e., L ≈ f = 60 cm).
For the first-order maximum, we have:
d sin θ = λ
0.0002 sin θ = 5 × 10^-7
sin θ ≈ 0.0025
θ ≈ 0.144 degrees
y/L ≈ 0.0025
y ≈ 0.15 cm
So the position of the first-order maximum is approximately 0.15 cm from the center of the screen.
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A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. Can you explain this answer?
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A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. 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 A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. Can you explain this answer?.
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. 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 A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10–5cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is :-Correct answer is '0.15'. Can you explain this answer?.
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