Question

The fringe width in a Youngs double slit interference pattern is $2.4\times 10^{-4}m$, when red light of wavelength $6400\mathring{A}$ is used. How much will it change, if blue light of wavelength $4000\mathring{A}$ is used?

• A. $9\times 10^{-4}m$
• B. $0.9\times 10^{-4}m$
• C. $4.5\times 10^{-4}m$
• D. $0.45\times 10^{-4}m$

Answer 1

Answer: (B)

Here, ${\beta }_1=2.4\times 10^{-4}m$
${\lambda }_1=6400\mathring{A},{\lambda }_2=4000\mathring{A}$
$\because \frac{{\beta }_2}{{\beta }_1}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{4000}{6400}=\frac{5}{8}$
or ${\beta }_2=\frac{5}{8}\times {\beta }_1=\frac{5}{8}\times 2.4\times 10^{-4}$
$=1.5\times 10^{-4}m$
Decrease in fringe width
$\Delta \beta ={\beta }_1-{\beta }_2=\left(2.4-1.5\right)\times 10^{-4}$
$=0.9\times 10^{-4}m$