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Solve the following.
Two identical rectangular slits 5 mm apart are illuminated with a monochromatic light of wavelength…
Two identical rectangular slits $5\text{ mm}$ apart are illuminated with a monochromatic light of wavelength $600\text{ nm}$. The screen is kept $1.2\text{ m}$ away from the slits.
(i)[1.5]
Calculate the distance between the $5^{\text{th}}$ bright fringe (band) on one side and the $3^{\text{rd}}$ bright fringe on the other side of the central bright band.
(ii)[1.5]
What will be the change in the interference pattern if the given light is replaced with monochromatic light of wavelength $500\text{ nm}$?
Answer
Answer (i)
Fringe width $\beta = \frac{\lambda D}{d} = \frac{600\times10^{-9}\times1.2}{5\times10^{-3}} = 1.44\times10^{-4}\text{ m}$. On opposite sides of the central band the 5th and 3rd bright fringes are at $5\beta$ and $3\beta$, so the distance is $8\beta = 8\times1.44\times10^{-4} = 1.152\times10^{-3}\text{ m}$.
Final answer: 1.152 mm
Answer (ii)
Fringe width $\beta \propto \lambda$, so it decreases from $0.144\text{ mm}$ to $0.12\text{ mm}$: the fringes become narrower and closer together (the pattern contracts).