Answer to Question #252931 in Optics for God

Question

Interference fringes are produced by monochromatic light falling normally on a wedge shaped film of cellophane of refractive index 1.4. If angle of wedge is 20 seconds of an arc and the distance between successive fringes is 0.25 cm, calculate the wavelength of light.

Answer 1

$\mu=1.4,\\\theta=20''=\dfrac{20\degree}{60\times 60}=\dfrac{1\degree}{180}=\dfrac{1}{180}\times \dfrac{\pi}{180}\ radian\\\beta=0.25\ cm\\\lambda=?$

We know that,

$\beta=\dfrac{\lambda}{2\mu\theta}\ \ or\ \ \lambda=2\beta\mu\theta\\\ \\\therefore\ \lambda=2\times 0.25\times 1.4\times \dfrac{\pi}{180\times 180}\\\ \\\lambda=6.787\times 10^{-5}cm= 6787 A\degree$