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Fresnel's Biprism - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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5 Questions around this concept.
Solve by difficulty
In a Fresnel's Biprism experiment Interference Fringes are observed with a biprism of refracting angle 2° and refractive index 1.5 on a screen which is 100m away from the source. If the distance between the source and the biprism is 20m and the Fringes width is 0.10 mm what is the wavelength of light ?
If prism angle , the distance between screen and prism (b)
, the distance between prism and source,
then in Fresnel biprism find the value of
(fringe width) :
How large can be the aperture opening to work with laws of ray optics using a monochromatic light of wavelength 450 nm to a distance of around 20 m?
How large can be aperture opening to work with laws of ray optics using a monochromatic light of wavelength 800 nm to a distance of around 20 m ?
Concepts Covered - 1
Fresnel's Biprism
- It is an optical device for producing interference of light Fresnel's biprism is made by joining base to base two thin prism of very small angle.
- When a monochromatic light source is kept in front of biprism two coherent virtual source S1 and S2 are produced.
- Interference fringes are found on the screen placed behind the biprism interference fringes are formed in the limited region which can be observed with the help eyepiece.
- Fringes are of equal width and its value is $\beta=\frac{\lambda D}{d}$
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- Let the separation between $\mathrm{S}_1$ and $\mathrm{S}_2$ be d and the distance of slits and the screen from the biprism be a and b respectively i.e. $D=(a+b)$. If the angle of the prism is $A$ and the refractive index is $\mu_{\text {then }} d=2 a(\mu-1) A$
$
\therefore \quad \lambda=\frac{\beta[2 a(\mu-1) A]}{(a+b)} \Rightarrow \beta=\frac{(a+b) \lambda}{2 a(\mu-1) A}
$
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