1. Tardigrade
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  4. A Fresnel biprism of angle 2° is illuminated by light of wavelength 6280 mathringA from a source which is 0.10 m away from it. What will be the width of the fringes (in mm) formed on a screen kept 0.9 m away from the biprism? (Take Refractive index of glass =1.5, π=3.14 )

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Q. A Fresnel biprism of angle $2^{\circ}$ is illuminated by light of wavelength $6280\,\mathring{A}$ from a source which is $0.10\, m$ away from it. What will be the width of the fringes (in mm) formed on a screen kept $0.9\, m$ away from the biprism?
(Take Refractive index of glass $=1.5, \pi=3.14$ )

Wave Optics

Solution:

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The deviation $\delta$ produced by each half of biprism $=(\mu-1) A$ where, $A$ is in radian.
Here,
$d =2(\mu-1) Ax$
$=2(1.5-1)\left(\frac{\pi}{180} \times 2^{\circ}\right) \times 0.1$
$d =\frac{\pi}{900} m$
$D =0.9+0.1=1\, m$
Now, $\beta=\frac{\lambda D }{ d }$
$\beta=\frac{\left(6280 \times 10^{-10}\right) \times 1}{\frac{\pi}{900}}$
$=1.8 \times 10^{-4} m =0.18\, mm$