Question

Calculate the wavelength of light used in an interference experiment from the following data: Fringe width $=0.03cm$. Distance between the slits and eyepiece through which the interference pattern is observed is $1m$. Distance between the images of the virtual source when a convex lens of focal length $16cm$ is used at a distance of $80cm$ from the eyepiece is $0.8cm$.

• A. $0.006\stackrel{\circ }{A}$
• B. $0.0006m$
• C. $600cm$
• D. $6000\stackrel{\circ }{A}$

Answer 1

The correct option is D $6000\stackrel{\circ }{A}$

Given: fringe with $\beta =0.03cm,D=1m=100cm$
Distance between images of the source $=0.8cm$
Image distance $v=80cm$
Object distance $=u$
Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{60}+\frac{1}{u}=\frac{1}{16}$
$\Rightarrow u=20cm$
Magnification, $m=\frac{v}{u}=\frac{80}{20}=4$
$Magnification=\frac{\text{distances between images of slits}}{\text{distance between slits}}$
$=\frac{0.8}{d}=\frac{0.8}{d}=4\Rightarrow d=0.2cm$
Fringe width $\beta =\frac{D\lambda }{d}$
or, $\beta =\frac{100\lambda }{0.2}=0.03\times {10}^{-2}$
Therefore, wavelength of light used $\lambda =6000\stackrel{\circ }{A}$.