Question:
A fringe width of $6 \mathrm{~mm}$ was produced for two slits separated by $1 \mathrm{~mm}$ apart. The screen is placed $10 \mathrm{~m}$ away. The wavelength of light used is ' $\mathrm{x}$ ' $\mathrm{nm}$. The value of ' $\mathrm{x}$ ' to the nearest integer is______
Solution:
$(600)$
$\beta=\frac{\lambda D}{d}$
$\lambda=\frac{\beta d}{D}$
$\lambda=\frac{6 \times 10^{-3} \times 10^{-3}}{10}$
$\lambda=6 \times 10^{-7} \mathrm{~m}=600 \times 10^{-9} \mathrm{~m}$
$\lambda=600 \mathrm{~nm}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.