In Lloyd's mirror experiment, a light wave emitted directly by the source $S$ (a narrow slit) interf

Question

In Lloyd's mirror experiment, a light wave emitted directly by the source $S$ (a narrow slit) interferes with the wave reflected from a mirror $M$. As a result, an interference fringe pattern is formed on the screen $Sc$. The source and the mirror screen are separated by a distance $l = 100 	ext{ cm}$. At a certain position of the source, the fringe width on the screen was equal to $\Delta x = 0.25 	ext{ mm}$, and after the source was moved away from the mirror plane by $\Delta h = 0.60 	ext{ mm}$, the fringe width decreased $\eta = 1.5$ times. Find the wavelength of light (in nm).

Professor PC Thomas Classes · Answer

In Lloyd's mirror, the interference is formed between the actual source and its virtual image. If the source is at height $h$ from the mirror, the distance between the sources is $d = 2h$. The fringe width is $\Delta x = \frac{\lambda l}{d} = \frac{\lambda l}{2h}$. When the source is moved to $h + \Delta h$, the new fringe width is $\Delta x' = \frac{\lambda l}{2(h + \Delta h)}$. We are given $\Delta x' = \frac{\Delta x}{\eta}$, so $\frac{\Delta x}{\eta} = \frac{\lambda l}{2h + 2\Delta h}$. From the first equation, $2h = \frac{\lambda l}{\Delta x}$. Substituting this in, we get $\frac{\Delta x}{\eta} = \frac{\lambda l}{\frac{\lambda l}{\Delta x} + 2\Delta h}$. Solving for $\lambda$ yields $\frac{1}{\eta} = \frac{\lambda l}{\lambda l + 2\Delta h \cdot \Delta x} \implies \lambda l + 2\Delta h \cdot \Delta x = \eta \lambda l \implies \lambda = \frac{2\Delta h \cdot \Delta x}{l(\eta - 1)}$. Substituting the given values: $\Delta h = 0.6 	imes 10^{-3} 	ext{ m}$, $\Delta x = 0.25 	imes 10^{-3} 	ext{ m}$, $l = 1 	ext{ m}$, $\eta = 1.5$. Thus, $\lambda = \frac{2(0.6 	imes 10^{-3})(0.25 	imes 10^{-3})}{1(1.5 - 1)} = \frac{0.3 	imes 10^{-6}}{0.5} = 0.6 	imes 10^{-6} 	ext{ m} = 600 	ext{ nm}$.