Question

In a Fresnel’s biprism experiment using Sodium light, fringe widths are obtained at a distance of 1m 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 1m 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.

Answer 1

The distance of the object can be obtained using the formula for magnification and then this value of distance of the object can be used to find the wavelength of the sodium source of light.

Formula Used: The wavelength of the source of monochromatic light can be obtained using the simple expression give below:
$\lambda = \dfrac{{\beta d}}{D}$
Here $\beta$is the fringe width.
$d$is the distance of the object
$D$is the distance of the slit

Step by step answer:
The fringe width is given in the question as $\beta = 195\mu m = 195 \times {10^{ - 6}}m$
Now we know that the magnification produced is a ratio between the size of image to that of the object and also the ratio between the distance of the image to that of the object. Thus,
$m = \dfrac{{{d_1}}}{d} = \dfrac{v}{u}$
According to the question, ${d_1} = 7cm = 0.007m$and $u = 30cm = 0.3m$also, $v = 1 - 0.3 = 0.7m$. Thus putting the respective values in the equation we get:
$\ d = \dfrac{{{d_1} \times u}}{v} \\\ \Rightarrow d = \dfrac{{0.007 \times 0.3}}{{0.7}} = 3 \times {10^{ - 3}}m \\\ \$
Now we have the distance between the slit is given as $D = 1m$. Now putting all respective values in the formula, we get:
$\lambda = \dfrac{{\beta d}}{D} = \dfrac{{195 \times {{10}^{ - 6}} \times 3 \times {{10}^{ - 3}}}}{1} = 585 \times {10^{ - 9}}m$
Now, we can either choose to keep the final answer in metres as that is the SI unit but it is better to change it into the respective unit that is nanometres (nm).
Now, we know that $1nm = {10^{ - 9}}m$
Thus the wavelength of the monochromatic sodium source will be $\lambda = 585nm$.

Note: The Fresnel biprism experiment is widely used to obtain the wavelength of the source of light. However, this can only be done if the source is monochromatic in nature, that is it only has a single wavelength and not a mixture of a range of different wavelengths.