The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavelength `6400 Å` is used. By how much will it change, if blue light of wavelength `4000 Å` is used ?A. `9 xx10^(-4)m`B. `0.9 xx10^(-4)m`C. `4.5 xx10^(-4)m`D. `0.45 xx10^(-4)m`

The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavel

1.	The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavelength `6400 Å` is used. By how much will it change, if blue light of wavelength `4000 Å` is used ?A. `9 xx10^(-4)m`B. `0.9 xx10^(-4)m`C. `4.5 xx10^(-4)m`D. `0.45 xx10^(-4)m`
Answer» Correct Answer - B Here, `beta_(1)=2.4 xx10^(-4)m,` ` lambda_(1)=6400 Ã…, lambda_(2)=4000 Ã…` ` :. (beta_(2))/(beta_(1))=(lambda_(2))/(lambda_(1))=(4000)/(6400)=(5)/(8)` or `beta_(2)=(5)/(8)xx beta_(1)=(5)/(8)xx2.4xx10^(-4)=1.5xx10^(-4)m` Decrease in fringe width `triangle beta=beta_(1)-beta_(2)=(2.4-1.5)xx10^(-4)=0.9xx10^(-4)m`

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The fringe width in YDSE is `2.4 xx 10^(-4)m`, when red light of wavelength `6400 Å` is used. By how much will it change, if blue light of wavelength `4000 Å` is used ?A. `9 xx10^(-4)m`B. `0.9 xx10^(-4)m`C. `4.5 xx10^(-4)m`D. `0.45 xx10^(-4)m`

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