Question

In a YDSE setup interference fringes are obtained by sodium light of wavelength $5890\mathring{A}$. On screen the fringes have an angular width $0.20^{\circ}$. Now the wavelength of light is changed and it is found that fringe width increases by $10 \%$, the new wavelength of incident light is:

• A. $5896\,\mathring{A}$
• B. $7321 \,\mathring{A}$
• C. $6300 \,\mathring{A}$
• D. $6479 \,\mathring{A}$

Answer 1

Answer: (D)

The angular width of interference fringes in YDSE setup for the two wavelengths are given as
$\theta_{1}=\frac{\lambda_{1}}{d} $ and $ \theta_{2}=\frac{\lambda_{2}}{d} $
$\Rightarrow \frac{\theta_{1}}{\theta_{2}}=\frac{\lambda_{1}}{\lambda_{2}} $
$\Rightarrow \lambda_{2}=\lambda_{1} \times \frac{\theta_{2}}{\theta_{1}}$
$\Rightarrow \lambda_{2}=5890 \times \frac{0.22}{0.20}=6479 \,\mathring{A}$