Answer to Question #240024 in Optics for Swayam

Question

 Interference fringes are produced with monochromatic light falling normally on a wedge￾shaped film of refractive index 1.4. The angle of the wedge is 10 seconds of an arc and 

the distance between the successive fringes is 0.5 cm. Calculate the wavelength of light 

used.

Answer 1

$\alpha =10'' \approx4.85*10^{-5}rad$

$n = 1.4\text{ refractive index}$

$b = 0.5cm = 5*10^{-3}m \text{ the distance between the successive fringes}$

$OPD -\text{optical path difference }$

$\lambda\text{ wavelength of light }$

$\text {Let } d\text{ the thickness of the film at the place of incidence of the beam}$

$\text{For a beam reflected from the upper surface:}$

$OPD_1 =-\frac{\lambda}{2};$

$\text{For a beam reflected from the inner surface of the film:}$

$OPD_2 =2dn$

$OPD =2dn- \frac{\lambda}{2}$

$\text{Minimum interference condition:}$

$OPD = k\lambda -\frac{\lambda}{2}; \ k \in Z$

$2dn = k\lambda$

$\text{the transition to the neighboring strip corresponds to a change in k per unit}$

$2(d+d')n= (k+1)\lambda$

$d'= \frac{\lambda }{2n}$

$\tan \alpha= \frac{d'}{b}$

$\lambda = 2nb\tan \alpha$

$\tan α ≈ α\text{ for small angles in radians}$

$\lambda = 2*1.4*5*10^{-3}*4.85*10^{-5}= 6.79*10^{-7}m$

$\text{Answer: }\lambda = 6.79*10^{-7}m$