Question

In ydse distance between slits and screen is 1.5 m when light of wavelength 500 nm is used then 2nd bright fringe is obtained on screen at a distance of 10 mm

Answer 1

Answer:

Young's Double Slit Experiment

Explanation:

GIVEN ;

★ Wavelength { lambda } = 500 nm

★ D { Distance between slit and screen } = 1.5m

★ n = No. of Fringes = 2 ( here )

★ Y { Displacement of Fringe } = 10 mm

★ d { Slit Width } = ????

«★» BY USING FORMULA ;

♥ Y = [ n × lambda × D ] / d ♠

» Y = [ 2 × 500 × 10^-9 × 1.5 ] / d

» 10 × 10-³ = [ 1000 × 10^-9 × 1.5 ] / d

» d = [ 10^-6 × 1.5 ] / 10-²

» d = 1.5 × 10-²

» d = 15 × 10-³ m OR 15 mm.

ANSWER :- SLIT Width for Wavelength if 500 nm is 15 Millimeters.