Question

In young's double slit experiment, the fringe width with light of wavelength ${\lambda }_1$ is ${\beta }_1$ and with light of wavelength ${\lambda }_2$ is ${\beta }_2$. Using light of wavelength ${\lambda }_1$ , the fringe width becomes ${\beta }_3$ if the entire apparatus is immersed in a transparent liquid of refractive index $\mu$. Then,

• A. =
• B. =
• C. =
• D. =

Answer 1

Answer: (B), (C)

The correct options are
B

=

C

=

${\beta }_1$ = $\frac{{\lambda }_{1}D}{d}$ ${\beta }_2$ = $\frac{{\lambda }_{2}D}{d}$

So, ${\beta }_2$= ${\beta }_1$$\frac{{\lambda }_2}{{\lambda }_1}$

$\frac{{\beta }_1}{{\beta }_2}$ = $\frac{{\lambda }_1}{{\lambda }_2}$ , now ${\lambda }_1$ = $\frac{{\lambda }_a}{\mu }$ = $\frac{{\lambda }_1}{\mu }$

Therefore ${\beta }_3$ = $\frac{{\beta }_1}{\mu }$