Fresnel biprism experiment – Physics – eSaral

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## Fresnel’s Biprism Experiment

1. It is an optical device to obtain two coherent sources by the refraction of lights.
2. The angle of biprism is $179^{\circ}$ & refracting angle is $\alpha=1 / 2^{\circ}$.

3. Distance between source & screen D = a + b. Distance between two coherent sources

$=\mathrm{d}=2 \mathrm{a}(\mu-1) \alpha$

Where a = distance between source & Biprism

b = distance between screen & Biprism

$\mu$ = refractive index of the material of the prism.

$\lambda=\frac{d \beta}{D}=\frac{2 a(\mu-1) \alpha \beta}{(a+b)}$

$=\frac{\sqrt{d_{1} d_{2}} \cdot \beta}{(a+b)}$

Note-

$\alpha$ is in radian $\alpha^{0}=\alpha \times \frac{3.14}{180}$ Suppose refracting angle & refractive index is not known then d can be calculated by a convex lens.

One convex lens whose focal length (f) and 4f < D. First convex lens is kept near biprism & $\mathrm{d}_{1}$ is calculated then it is kept near eyepiece & $\mathrm{d}_{2}$ is calculated. $\mathrm{d}=\sqrt{\mathrm{d}_{1} \mathrm{~d}_{2}}$

Application :

With the help of this experiment the wavelength of monochromatic light, the thickness of thin films, and their refractive index & distance between apparent coherent sources can be determined. When Fresnel’s arrangement is immersed in water

(a) Effect on d

$\mathrm{d}_{\text {water }}<\mathrm{d}_{\mathrm{air}}$. Thus when the Fresnel’s biprism experiment is immersed in water, then the separation between the two virtual sources decreases but in young’s double-slit experiment it does not change.

(b) In young’s double-slit experiment $\beta$ decreases and in Fresnel’s biprism experiment $\beta$ increases.

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