Question

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

• A. $0.0006\,\mathring{A}$
• B. $0.0006\, cm$
• C. $600\, cm$
• D. $6000\,\mathring{A}$

Answer 1

Answer: (D)

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