in a biprism experiment, light of wavelength 5200 A is used to get an interference pattern on the screen.the fringe width changes by 1.3mm when the screen is moved towards biprisim by 50cm.find the distance between two virtual images of the slit........
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formula of fringe width is given by
β = λD/d
Here β is fringe width , λ is wavelength of wave , D is the distance between source and screen , d is the distance between two virtual image of the slit.
Here, λ = 5200 A° = 5.2 × 10⁻⁷m
D = 50 cm = 0.5 m
β = 1.3 mm = 1.3 × 10⁻³ m
d = ? { distance between virtual image of the slit }
1.3 × 10⁻³ = 5.2 × 10⁻⁷ × 0.5/d
d = 4 × 10⁻⁴ × 0.5 = 2 × 10⁻⁴ = 0.2 × 10⁻³ m = 0.2mm
Hence, answer should be 0.2 mm
β = λD/d
Here β is fringe width , λ is wavelength of wave , D is the distance between source and screen , d is the distance between two virtual image of the slit.
Here, λ = 5200 A° = 5.2 × 10⁻⁷m
D = 50 cm = 0.5 m
β = 1.3 mm = 1.3 × 10⁻³ m
d = ? { distance between virtual image of the slit }
1.3 × 10⁻³ = 5.2 × 10⁻⁷ × 0.5/d
d = 4 × 10⁻⁴ × 0.5 = 2 × 10⁻⁴ = 0.2 × 10⁻³ m = 0.2mm
Hence, answer should be 0.2 mm
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