Question

Describe the biprism experiment to find the wavelength of the monochromatic light. Draw the necessary ray diagram.
The width of plane incident wavefront is found to be doubled on refraction in denser medium. If it makes an angle of `65^(@)` with the normal, calculate the refractive index for the denser medium.

Answer 1

Biprism experiment to find the wavelength of the monochromatic light :
To find the wavelength of the monochromatic light through Biprism experiment, an optical bench is used. The length of optical bench is about one and half meter long and scale is marked along its length. Four adjustable stands carrying the slit(S), biprism (B), lens (L) and the micrometer eyepiece (E) are mounted on optical bench. The slit and the biprism can be rotated about horizontal axis.
The slit, biprism and eyepiece are required to be set at same height by keeping their centres in same line.The slit is made narrow and is illuminated by monochromatic light (viz. sodium vapour lamp). The slit is kept vertical and the stand carrying slit is kept close to the stand carrying biprism. Now, rotate the biprism such that its refracting edge becomes parallel to slit and the interference pattern consisting of alternate bright and dark bands appears in the field view of the eyepiece. After this the wavelength of light can easily be determined by using equation,
`X=(lambda D)/(d)`
`or lambda=(Xd)/(D)`

B = Biprism
S = Slit
L = Convex lens
E = Eyepiece
Numerical :
Given: `angle i = 65^(@)`
From figure, `cos i=(AB)/(AD) and cos r=(CD)/(AD)`

`therefore (cos r)/(cos i) =(CD)/(AB)=2`(given)
`therefore cos r=2xx cos i`
`cos r=2xx cos 65^(@)`
`=2xx0.423`
`=0.8425`
`therefore r=cos^(-1)(0.845)`
`=32.328^(@)`
Now, `" "_(r)mu_(d)=(sin i)/(sin r)`
`=(sin65^(@))/(sin32.328^(@))`
`=(0.906)/(0.535)`
`" "_(r)mu_(d)=1.693`
`therefore` Refractive index for the denser medium is 1.693.