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 ?

Options

A. `9 xx10^(-4)m`
B. `0.9 xx10^(-4)m`
C. `4.5 xx10^(-4)m`
D. `0.45 xx10^(-4)m`

Answer

B

Explanation

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`