Question

In an experiment with a biprism, the readings on the optical bench of the position of the eyepiece and the two positions of the lens were respectively $1.0m$, $0.67m$ and $34.00cm$. The distances between the two images for the two positions of the lens were respectively $0.3000mm$ and $1.2000mm$ and the width of $10$ fringes was $9.720mm$. Assuming that there is no index error in any case, calculate (i) the distance between the focal plane of interfering sources, and (ii) the wavelength of the light used.

Answer 1

The distance between the eyepiece and the second position of the lens is given by

$V_1=100-67=33cm=4$

Hence, the reading of the slit on the bench is

$=$ first position of the lens$-4$

$=34-44=1cm$

Hence the distance between the focal plane of the eye piece and the plane of interfering sources is

$D=100-1$

$=99cm=0.99m$

$\left(b\right)$ The separation of the two intersecting sources is

$2d=\sqrt{d_1\times d_2}$

Here, $d_1=1.2㎜$

$0.12㎝$

and $d_2=0.3㎜=0.03㎝$

$\therefore 2d=\sqrt{0.12\times 0.03}$

$=0.06㎝$

Width of $10$ fringes was $9.720㎜$

$\therefore 10w=9.720㎜$

$=0.972㎝$

or $w=\frac{0.972}{10}=0.0972㎝$

$\therefore$ The wavelength of light

$\lambda =\frac{2d}{D}w$

$w=\frac{0.06\times 0.0972}{99}$

$w=5891\times {10}^{-8}㎝$

$w=589.1㎚$