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In a biprism expt., by using light of wavelength 5000 Å, 5 mm wide fringes are obtained on a screen 1.0 m away from coherent sources. The separation between two coherent sources is
- a)1.0 mm
- b)0.1 mm
- c)0.05 mm
- d)0.01 mm
Correct answer is option 'B'. Can you explain this answer?
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In a biprism expt., by using light of wavelength 5000 , 5 mm wide frin...
Concept: In biprism experiment, the fringe width is given by the formula:
w = λL/d
where w is the fringe width, λ is the wavelength of light used, L is the distance between the biprism and the screen, and d is the distance between the two coherent sources.
Given:
λ = 5000 Å (angstroms)
w = 5 mm
L = 1.0 m
Calculation:
1 mm = 1000 µm
1 µm = 10^-6 m
1 Å = 10^-10 m
Converting the given values to SI units:
λ = 5000 × 10^-10 m
w = 5 × 10^-3 m
L = 1.0 m
Substituting these values in the formula, we get:
w = λL/d
d = λL/w
d = (5000 × 10^-10 × 1)/5 × 10^-3
d = 10^-7 m
d = 0.1 µm
Therefore, the separation between the two coherent sources is 0.1 µm.
Answer: Option (b) 0.1 mm
w = λL/d
where w is the fringe width, λ is the wavelength of light used, L is the distance between the biprism and the screen, and d is the distance between the two coherent sources.
Given:
λ = 5000 Å (angstroms)
w = 5 mm
L = 1.0 m
Calculation:
1 mm = 1000 µm
1 µm = 10^-6 m
1 Å = 10^-10 m
Converting the given values to SI units:
λ = 5000 × 10^-10 m
w = 5 × 10^-3 m
L = 1.0 m
Substituting these values in the formula, we get:
w = λL/d
d = λL/w
d = (5000 × 10^-10 × 1)/5 × 10^-3
d = 10^-7 m
d = 0.1 µm
Therefore, the separation between the two coherent sources is 0.1 µm.
Answer: Option (b) 0.1 mm
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In a biprism expt., by using light of wavelength 5000 , 5 mm wide fringes are obtained on a screen 1.0 m away from coherent sources. The separation between two coherent sources isa)1.0 mmb)0.1 mmc)0.05 mmd)0.01 mmCorrect answer is option 'B'. Can you explain this answer?
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