Question

A double-slit arrangement produces interference fringes for sodium light $(\lambda =589nm)$ that hae an angular separation of $3.50\times 0^{-3}rad$. For what wavelength would the angular separation be $10$% greater?

• A. $527nm$
• B. $648nm$
• C. $722nm$
• D. $449nm$

Answer 1

Answer: (B)

$648nm$

Angular separation $\theta =\frac{n\lambda }{d}\Rightarrow \theta \propto \lambda$

${\lambda }_2={\lambda }_1\frac{{\theta }_2}{{\theta }_1}=\left(589\right)\frac{(3.5\times {10}^{-3}+0.1\times 3.5\times {10}^{-3})}{3.5\times {10}^{-3}}=647.9\approx 648nm$