Question

In a Lloyd's single mirror experiment, the distance between the slit source and its image is $5mm$. The screen is at a distance of $1\text{m}$ from the source. The fringe width is observed to be $0.1mm$. Then the wavelength of light used is

• A. $5460\stackrel{\circ }{A}$
• B. $5000\stackrel{\circ }{A}$
• C. $6460\stackrel{\circ }{A}$
• D. $4000\stackrel{\circ }{A}$

Answer 1

The correct option is B $5000\stackrel{\circ }{A}$


In Lloyd's single mirror experiment, the image formed by the mirror $I$ acts as the second source at a phase difference of $\pi$ with $S$ (due to reflection). Therefore, interference is observed only in the limited region.
Fringe width is given by the same formula.
$\beta =\frac{\lambda D}{d}$
$\Rightarrow \lambda =\frac{\beta d}{D}=\frac{0.1\times {10}^{-3}\times 1}{5\times {10}^{-3}}$
$=5\times {10}^{-7}m=5000\stackrel{\circ }{A}$