Solveeit Logo
Login

Question

Physics Question on Wave optics

The fringe width in a Youngs double slit interference pattern is 2.4×10−4 m2.4 \times 10^{-4}\, m, when red light of wavelength 6400?6400 ? is used. How much will it change, if blue light of wavelength 4000?4000 ? is used?

A

9×10−4 m9 \times 10^{-4}\,m

B

0.9×10−4 m0.9 \times 10^{-4}\,m

C

4.5×10−4 m4.5 \times 10^{-4}\,m

D

0.45×10−4 m0.45 \times 10^{-4}\,m

Answer

0.9×10−4 m0.9 \times 10^{-4}\,m

Explanation

Solution

Here, β1=2.4×10−4m\beta_{1} = 2.4 \times 10^{-4} m λ1=6400?,λ2=4000?\lambda_{1} = 6400 ?, \lambda_{2} = 4000 ? ∵β2β1=λ2λ1=40006400=58\because\frac{\beta_{2}}{\beta_{1}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{4000}{6400} = \frac{5}{8} or β2=58×β1=58×2.4×10−4\beta_{2} = \frac{5}{8}\times\beta_{1} = \frac{5}{8}\times 2.4\times 10^{-4} =1.5×10−4m= 1.5 \times 10^{-4} m Decrease in fringe width Δβ=β1−β2=(2.4−1.5)×10−4 \Delta\beta = \beta_{1} -\beta_{2} = \left(2.4 - 1.5 \right)\times10^{-4} =0.9×10−4m = 0.9 \times 10^{-4} m