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)\times10^{-4}$
$ = 0.9 \times 10^{-4} m$