Question

In a Fresnel's biprism experiment using sodium light, fringe widths of $195$$\mu$m are obtained at a distance of $1.0m$ from the slit . A convex lens is then put between the observer and prism so as to give the image of the source at a distance of $1.0m$ from the slit . The distance apart of the images is found to be $0.7cm$, the lens being $30cm$ from the slit . Calculate the wavelength of the sodium light used.

• A. $595nm$
• B. $540nm$
• C. $585nm$
• D. $590nm$

Answer 1

The correct option is C $585nm$

Here fringe width $\beta =195\mu m=195\times {10}^{-6}m$
When the lens is at a distance of 0.3 m from the slit, the distance between the two coherent images becomes 0.007 m.
Now the magnification is given as
$m=\frac{\text{Size of the image}}{\text{Size of the object}}=\frac{\text{Distance of the image}}{\text{Distance of the object}}$
Now the distance between the images $=d_1=0.007m$
Distance between the objects $d$
Distance of the image from the lens is $1-0.3=0.7m$
Distance of the object from the lens $u=0.3m$
Thus
$\frac{d_1}{d}=\frac{v}{u}$
or
$d=\frac{d_{1}u}{v}=\frac{0.007\times 0.3}{0.7}=3\times {10}^{-3}m$
Now
$\lambda =\frac{\beta d}{D}=\frac{195\times {10}^{-6}\times 3\times {10}^{-3}}{1}=585nm$