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A beam of monochromatic light of wavelength 5.82 × 10^-7m falls normally on a glass wedge with the wedge angle of 20 seconds of an arc. If the refractive index of glass is 1.5, find the number of dark fringes per cm of the wedge length.

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5 per cm

3 per cm

6 per cm

4 per cm

answer is C.

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Detailed Solution

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At P, path difference for near normal incidence $= 2{\mkern 1mu} {\kern 1pt} {\mkern 1mu} {\kern 1pt} X\,PQ = 2\,x\theta$

So number of dark fringes on a length x, is given by $n= 2\,\mu \,x\,\frac{\theta}{\lambda}$

$\therefore$ Number of dark frings formed per unit length

$=\frac{2\mu\theta}{\lambda}=\frac{2\times1.5\times20\times5}{5.82\times180\times180}\times10^5\, per\,\, cm$

$=5 \,per\, cm.$

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