Question

In Lloyd's mirror experiment, a light wave emitted directly by the source $S\left(\text{narrow slit}\right)$ interferes with the wave reflected from a mirror $M$. As a result, an interference fringe pattern is formed on the screen which is $1\text{m}$ far from mirror. The source and the mirror are separated by a distance $l=100\text{cm}$. At a certain position of the source, the fringe width on the screen was equal to $\beta =0.25\text{mm}$ and after the source was moved away from the mirror plane by $\Delta h=0.60\text{mm}$ the fringe width decreased by $1.5\text{times}$. Find the wavelength of light from the source.

• A. $0.2\mu \text{m}$
• B. $0.6\mu \text{m}$
• C. $1\mu \text{m}$
• D. $0.4\mu \text{m}$

Answer 1

The correct option is B $0.6\mu \text{m}$


In Lloyd's experiment, the actual source and the image of the source acts as coherent sources to form an interference pattern.

Fringe width is given by,

$\beta =\frac{\lambda D}{d}$

When the source is moved away from the mirror by $0.6\text{mm}$, image of the source also moves away from the mirror by $0.6\text{mm}$.

Hence, $d=$ distance between the slits is increased by $1.2\text{mm}$

$\Rightarrow {\beta }^{\prime }=\frac{\lambda D}{d+1.2\text{mm}}$

Given that,
${\beta }^{\prime }=\frac{\beta }{1.5}$

$\Rightarrow \frac{1}{d+1.2\text{mm}}=\frac{1}{1.5d}$

$\Rightarrow d=2.4\text{mm}$

Given that, $\beta =0.25\text{mm}$

$\Rightarrow \lambda =\frac{0.25\times {10}^{-3}\times 2.4\times {10}^{-3}}{1}$

$\Rightarrow \lambda =0.6\mu \text{m}$

Hence, option $\left(C\right)$ is correct.