Question

fringe width in Young's slit experiment of wave length 7200A⁰ is 0.10mm. find fringe width when distance between source and screen is double and distance of separation is halved when wavelength is wave length is 4500A⁰?​

Answer 1

Answer:

♐ FRINGE WIDTH ( ẞ² ) = 0.06 mm ♐

Explanation:

★ WAVE OPTICS ★

★ GIVEN ;

» Fringe Width ( ẞ¹ ) = 0.10 mm

» Wavelength ( Lamda 1 ) = 7200 Å

» Wavelength ( lambda 2 ) = 4500 Å

★ » Fringe Width ( ẞ² ) = ???

[ ⬜→ EVEN DOUBLING DISTANCE BETWEEN SLITS AND SCREEN WON'T AFFECT " FRINGE WIDTH ". ←⬜ ]

________ [ BY USING FORMULA ] ______

[ $( \frac{ \beta 1}{ \beta 2} ) \: = ( \: \frac{lambda \: 1}{lambda \: 2 \: } )$ ]

★ [ 0.10 × 10-³ / ẞ² ] = [ 7200 Å / 4500 Å ]

» [ 10-⁴ / ẞ² ] = [ 7200 / 4500 ]

» ẞ² × 7200 = 4500 × 10-⁴

» ẞ² = [ 4500 × 10-⁴ / 7200 ]

» ẞ² = [ 0.625 × 10-⁴ ]

♒» ẞ² = 62.5 μm / micro metre « ♒

——————— [ OR ] ———————

⏭ ẞ² = 0.06 mm / Millimeters ⏮

________________________________________

ANSWER :- FRINGE WIDTH OF WAVELENGTH 4500 Å IS ;

♣ ẞ = 0.06 mm ♥

⏩ NOTE :- FRINGE WIDTH IS ALWAYS MEASURED IN MILLIMETERS AND FRINGE WIDTH IS PROPORTIONAL TO WAVELENGTH . ⏪