While conducting Young's double slit experiment it was noted that fringe width with light of wavelength 6000 mathringA was 3 mm . If the experiment is repeated but this time wavelength of light used is 4000 mathringA , what will be the new fringe width?

Question

While conducting Young's double slit experiment it was noted that fringe width with light of wavelength 6000 mathringA was 3 mm . If the experiment is repeated but this time wavelength of light used is 4000 mathringA , what will be the new fringe width? (A) 3 mm (B) 1 mm (C) 2 mm (D) 4 mm

Tardigrade Admin · Accepted Answer

fringe width β =(λ D/d) Where λ is the wavelength of light, D is distance between slits and the screen, d is distance between the two slits. As D and d remain the same, β propto λ Or (β '/β )=(λ '/λ ) or β '=(λ ' β /λ ) Substituting the given values, we get β '=(4000 â« × 3 m m/6000 â«)=2 mm

Q. While conducting Young's double slit experiment it was noted that fringe width with light of wavelength $6000 \overset{˚}{A}$ was $3 mm$ . If the experiment is repeated but this time wavelength of light used is $4000 \overset{˚}{A}$ , what will be the new fringe width?

$3 mm$

$1 mm$

$2 mm$

$4 mm$

Q. While conducting Young's double slit experiment it was noted that fringe width with light of wavelength 6000A˚ was 3mm . If the experiment is repeated but this time wavelength of light used is 4000A˚ , what will be the new fringe width?

3mm

1mm

2mm

4mm

fringe width β=dλD​ Where λ is the wavelength of light, D is distance between slits and the screen, d is distance between the two slits. As D and d remain the same, β∝λ Or ββ′​=λλ′​ or β′=λλ′β​ Substituting the given values, we get β′=6000Å4000Å×3mm​=2mm

Q. While conducting Young's double slit experiment it was noted that fringe width with light of wavelength $6000 \overset{˚}{A}$ was $3 mm$ . If the experiment is repeated but this time wavelength of light used is $4000 \overset{˚}{A}$ , what will be the new fringe width?

$3 mm$

$1 mm$

$2 mm$

$4 mm$