Interference fringes from sodium light \( (\lambda=5890\,\mathrm{\mathring{A}}) \) in a double-slit experiment have an angular width of \( 0.20^\circ \). To increase the fringe width by 10%, the wavelength of light used should be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 6479\,\mathrm{\mathring{A}} \)
In the double-slit experiment the angular width of a fringe is \( \theta=\lambda/d \), and the fringe width on a fixed screen is \( \beta=\lambda D/d \). For the same slits and screen, both are proportional to the wavelength, so the given angular width \( 0.20^\circ \) is not needed.
A 10% larger fringe width needs a 10% larger wavelength:
\( \begin{aligned} \lambda' &= 1.1\,\lambda \\ &= 1.1\times5890\,\mathrm{\mathring{A}} \\ &= 6479\,\mathrm{\mathring{A}} \end{aligned} \)