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A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]
- a)0°
- b)15°
- c)30°
- d)50°
Correct answer is option 'C'. Can you explain this answer?
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A parallel beam of monochromatic light ofwavelength 5000Å is inc...
For first minimum, a sin θ = nλ = 1λ
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A parallel beam of monochromatic light ofwavelength 5000Å is inc...
Å is incident on a diffraction grating with 500 lines per mm. The diffraction pattern is observed on a screen placed at a distance of 1 m from the grating.
(a) What is the distance between two consecutive bright fringes on the screen?
(b) What is the angular separation between the first-order and second-order maxima?
(c) How many bright fringes will be observed on the screen if the width of the grating is 4 cm?
(d) What is the width of the central maximum on the screen?
Solution:
(a) The distance between two consecutive bright fringes can be calculated using the formula:
d sinθ = mλ
where d is the distance between the adjacent slits in the grating, θ is the angle between the incident beam and the direction of the mth maximum, m is the order of maximum, and λ is the wavelength of the incident light.
In this case, d = 1/500 mm = 2 × 10^-3 cm, m = 1, and λ = 5000 Å = 500 nm. Therefore, we have:
sinθ = mλ/d = 500 × 10^-9 / (2 × 10^-3) = 0.25
θ = sin^-1(0.25) = 14.48°
The distance between two consecutive bright fringes can be calculated using the formula:
y = D tanθ
where y is the distance between two consecutive bright fringes on the screen, and D is the distance between the grating and the screen.
In this case, D = 1 m. Therefore, we have:
y = D tanθ = 1 × tan(14.48°) = 0.25 m
Therefore, the distance between two consecutive bright fringes is 0.25 m.
(b) The angular separation between the first-order and second-order maxima can be calculated using the same formula as above:
θ2 - θ1 = sin^-1(mλ/d) - sin^-1((m-1)λ/d)
where m = 1 for the first-order maximum and m = 2 for the second-order maximum.
In this case, d = 1/500 mm = 2 × 10^-3 cm and λ = 5000 Å = 500 nm. Therefore, we have:
θ2 - θ1 = sin^-1(500 × 10^-9 / (2 × 10^-3)) - sin^-1(2 × 500 × 10^-9 / (2 × 10^-3))
θ2 - θ1 = 14.48° - 28.96° = -14.48°
Therefore, the angular separation between the first-order and second-order maxima is 14.48°.
(c) The number of bright fringes that will be observed on the screen can be calculated using the formula:
N = w/d
where N is the number of bright fringes, w is the width of the grating, and d is the distance between adjacent slits in the grating.
In this case, w = 4 cm and d = 1/500 mm = 2 × 10^-3 cm. Therefore, we have:
N = w/d = 4 × 10^-2 / (2 × 10^-3) = 20
(a) What is the distance between two consecutive bright fringes on the screen?
(b) What is the angular separation between the first-order and second-order maxima?
(c) How many bright fringes will be observed on the screen if the width of the grating is 4 cm?
(d) What is the width of the central maximum on the screen?
Solution:
(a) The distance between two consecutive bright fringes can be calculated using the formula:
d sinθ = mλ
where d is the distance between the adjacent slits in the grating, θ is the angle between the incident beam and the direction of the mth maximum, m is the order of maximum, and λ is the wavelength of the incident light.
In this case, d = 1/500 mm = 2 × 10^-3 cm, m = 1, and λ = 5000 Å = 500 nm. Therefore, we have:
sinθ = mλ/d = 500 × 10^-9 / (2 × 10^-3) = 0.25
θ = sin^-1(0.25) = 14.48°
The distance between two consecutive bright fringes can be calculated using the formula:
y = D tanθ
where y is the distance between two consecutive bright fringes on the screen, and D is the distance between the grating and the screen.
In this case, D = 1 m. Therefore, we have:
y = D tanθ = 1 × tan(14.48°) = 0.25 m
Therefore, the distance between two consecutive bright fringes is 0.25 m.
(b) The angular separation between the first-order and second-order maxima can be calculated using the same formula as above:
θ2 - θ1 = sin^-1(mλ/d) - sin^-1((m-1)λ/d)
where m = 1 for the first-order maximum and m = 2 for the second-order maximum.
In this case, d = 1/500 mm = 2 × 10^-3 cm and λ = 5000 Å = 500 nm. Therefore, we have:
θ2 - θ1 = sin^-1(500 × 10^-9 / (2 × 10^-3)) - sin^-1(2 × 500 × 10^-9 / (2 × 10^-3))
θ2 - θ1 = 14.48° - 28.96° = -14.48°
Therefore, the angular separation between the first-order and second-order maxima is 14.48°.
(c) The number of bright fringes that will be observed on the screen can be calculated using the formula:
N = w/d
where N is the number of bright fringes, w is the width of the grating, and d is the distance between adjacent slits in the grating.
In this case, w = 4 cm and d = 1/500 mm = 2 × 10^-3 cm. Therefore, we have:
N = w/d = 4 × 10^-2 / (2 × 10^-3) = 20
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A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer?
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A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer?.
A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to [1993]a)0°b)15°c)30°d)50°Correct answer is option 'C'. Can you explain this answer?.
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