Q 12-10-026JEE MainMedium
In Young's experiment with slit separation $1$ mm, the fringe width is $0.6$ mm. When the screen is moved $50$ cm farther from the slits, the fringe width becomes $0.9$ mm. Find the wavelength of the light in nm.
Numerical value type. Enter your answer.
Answer: 600
$\beta = \dfrac{\lambda D}{d}$, so the change in fringe width is $\Delta\beta = \dfrac{\lambda\,\Delta D}{d}$:
$$\lambda = \frac{\Delta\beta\,d}{\Delta D} = \frac{0.3 \times 10^{-3} \times 10^{-3}}{0.5} = 6 \times 10^{-7}\ \text{m} = 600\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics