Question

25. In young's double slit experiment light of wavelength 500 nm is used to obtain fringe
width 0.5 mm. If whole apparatus is immersed in water with refractive index 1.33.find
the new wavelength of light required so as to keep fringe width same.​ jaldi its urgent

Answer 1

Answer:

★ NEW WAVELENGTH = 376 nano metres. [ 376 × 10^-9 m ]

SO ; [ ẞ AIR = ẞ MEDIUM ]

Explanation:

★ FRINGE WIDTH ★

★ GIVEN ;

» Original Wavelength ( lambda¹ ) = 500 nm

» Fringe Width ( ẞ ) = 0.5 mm

» Refractive Index of Medium in which Set up is immersed ( μ ) = 1.33 [ Water ]

★ NEW WAVELENGTH ( LAMBDA² ) = ???

[ FRINGES WILL BE SAME IF NEW WAVELENGTH CORRESPONDS THE MEDIUM ]

_____ [ BY USING FORMULA ] ______

$[ lambda \: medium \: = \: \frac{1}{u} \: \times \: lambda \: air \: ]$

★ Lambda² = [ Lambda¹ / μ ]

» Lambda² = [ 500 × 10^-9 m / 1.33 ]

» Lambda² = [ 376 × 10^-9 m ]

★ NEW WAVELENGTH [ LAMBDA² ] = 376 × 10^-9 m [ OR ]

376 nm .

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★ [ ẞ AIR = ẞ MEDIUM ]

WHEN ; WAVELENGTH IN MEDIUM CORRESPONDS TO 376 nm.

SO ; FRINGE WIDTH IN WATER BECOMES 0.5 mm [ SAME AS AIR ] .

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