Question

Calculate the wavelength of light used in an interference experiment from the following data: Fringe width $=0.03cm$. Distance between 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. $6000\mathring{A}$
• B. $0.00006\mathring{A}$
• C. $6000cm$
• D. $0.00006m$

Answer 1

The correct option is A $6000\mathring{A}$

In the experiment,

$d=0.8cm,D=1m,\beta =0.03cm$

But this 'd' would be different from that used to produce interference pattern since the convex lens would cause a magnification changing the effective distance between the point sources.

Magnification of the lens=$\frac{v}{u}=\frac{0.2}{0.8}=\frac{1}{4}$

Thus the $d_{effective}$ is $\frac{d}{4}$

Thus $\beta =\frac{4D\lambda }{d}$

$\lambda =\frac{\beta d}{4D}=6000A^{\circ }$