Answer :
Answer:B
Explanation:
Given
Wavelength of light [tex]\lambda =590\ nm[/tex]
Screen distance [tex]L=80\ cm[/tex]
First fringe is at a distance [tex]y_1=1.9\ cm[/tex]
No of lines per mm is given by N
[tex]N=\frac{1}{d}[/tex]
where d=slit width
From N-slits Experiment
[tex]\sin \theta _m=\frac{m\lambda }{d}[/tex]
[tex]d=\frac{m\lambda }{\sin \theta _m}-----1[/tex]
Position of bright fringe is given by
[tex]y=\tan \theta _m\cdot L[/tex]
[tex]\tan \theta _m=\frac{y}{L}[/tex]
[tex]\theta _m=\tan^{-1}(\frac{y}{L})[/tex]
Put the value of [tex]\theta _m[/tex] in eq. 1
[tex]d=\frac{m\lambda }{\sin (\tan^{-1}(\frac{y}{L}))}[/tex]
Therefore [tex]N=d^{-1}[/tex]
[tex]N=\frac{\sin (\tan^{-1}(\frac{y}{L}))}{m\lambda }[/tex]
for [tex]m=1[/tex]
[tex]N=\frac{\sin (\tan^{-1}(\frac{1.9\times 10^{-2}}{0.8}))}{1\times 590\times 10^{-9}}[/tex]
[tex]N=40243\ line/m[/tex]
[tex]N=40\ line/mm[/tex]