Question

The fringes width in a Young’s double slit interference pattern is 2.4×10^-4 m, when red light of wavelength 6400 Å is used. How much will it change, if blue light os wavelength 400Å is used?

• A. 9×10^-4 m
• B. 0.9 ×10^-4 m
• C. 4.5× 10^-4 m
• D. 0.45 × 10^-4 m

Answer 1

Answer: (B)

0.9 ×10^-4 m

Answer 2

: Here, $\beta_{1} = 2.4 \times 10^{- 4}m$

$\lambda_{1} = 6400Å,\lambda_{2} = 4000Å$

$\because\frac{\beta_{2}}{\beta_{1}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{4000}{6400} = \frac{5}{8}$

Or $\beta_{2} = \frac{5}{8} \times \beta_{1} = \frac{5}{8} \times 2.4 \times 10^{- 4} = 1.5 \times 10^{- 4}m$

Decreases in fringe width

$\Delta\beta = \beta_{1} - \beta_{2} = (2.4 - 1.5) \times 10^{- 4} = 0.9 \times 10^{- 4}m$