Question

The fringe width of interference fringes for red colour is $\beta$. The fringe width for violet colour will be nearly :
$({\lambda }_{Red}=620-740nm,{\lambda }_{Violet}=380-450nm)$

• A. $\beta$
• B. 2$\beta$
• C. $\frac{\beta }{4}$
• D. $\frac{\beta }{2}$

Answer 1

The correct option is D $\frac{\beta }{2}$

${\lambda }_{red}=620-740nm$
${\lambda }_{violet}=380-450nm$
$So,{\lambda }_{red}\simeq {\lambda }_{violet}\times 2$
$or\frac{{\lambda }_{red}}{{\lambda }_{violet}}=2$
Fringe width $=\frac{D\lambda }{d}$
So, $\beta \propto \lambda$
Thus we get
$\frac{{\lambda }_{red}}{{\lambda }_{violet}}=\frac{{\beta }_{red}}{{\beta }_{violet}}=2$
or
${\beta }_{violet}=\frac{{\beta }_{red}}{2}$