Question

In young's double slit experiment the fringe width is 4mm, if the experiment refractive index 4/3 the fringe width becomes (in mm)

• A. 3
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 3

$\begin{array}{c}\beta =\frac{\lambda D}{d}\\ \mu =\frac{C}{V}=\frac{C}{f\lambda }\\ \Rightarrow 1=\frac{C}{f\lambda }\\ \&\frac{4}{3}=\frac{C}{f\lambda }\\ \Rightarrow \frac{4}{3}=\frac{\lambda }{{\lambda }^{\prime }}=\frac{\beta }{{\beta }^{\prime }}\\ \Rightarrow {\beta }^{\prime }=\frac{3}{4}\beta =3mm\\ Hence,\\ optionAiscorrectanswer.\end{array}$