Question

If the whole bi-prism experiment is immersed in water then the fringe width becomes, if the refractive indices of bi-prism material and water are $1.5$ and $1.33$ respectively,

• A. $3$ times
• B. $\frac{3}{4}$ times
• C. $\frac{4}{3}$ times
• D. $\frac{1}{3}$ times

Answer 1

Answer: (A)

$3$ times

Separation between the coherent sources when the entire setup is in air $d_{ga}=2a({\mu }_g-{\mu }_a)A$

Separation between the coherent sources when the entire setup is put inside water $d_{gw}=2a({\mu }_g-{\mu }_w)A$

where $a=$distance between the single slit and the biprism, A is the prism angle and ${\mu }_g$ is the redfractive index of biprism.

Fringe width ${\beta }_{ga}=\frac{D\lambda }{d_{ga}}$

Fringe width ${\beta }_{gw}=\frac{D\lambda }{d_{gw}}$

$\therefore \frac{{\beta }_{ga}}{{\beta }_{gw}}=\frac{d_{gw}}{d_{gw}}=\frac{{\mu }_g-{\mu }_w}{{\mu }_g-{\mu }_a}=\frac{1.5-1.33}{1.5-1}=\frac{0.17}{0.5}=\frac{0.17}{0.5}=\frac{1}{3}$

Hence, the fringe width increases 3 times.