Question #240024

 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.


1
Expert's answer
2021-09-24T09:26:57-0400

α=104.85105rad\alpha =10'' \approx4.85*10^{-5}rad

n=1.4 refractive indexn = 1.4\text{ refractive index}

b=0.5cm=5103m the distance between the successive fringesb = 0.5cm = 5*10^{-3}m \text{ the distance between the successive fringes}

OPDoptical path difference OPD -\text{optical path difference }

λ wavelength of light \lambda\text{ wavelength of light }

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

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

OPD1=λ2;OPD_1 =-\frac{\lambda}{2};

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

OPD2=2dnOPD_2 =2dn

OPD=2dnλ2OPD =2dn- \frac{\lambda}{2}

Minimum interference condition:\text{Minimum interference condition:}

OPD=kλλ2; kZOPD = k\lambda -\frac{\lambda}{2}; \ k \in Z

2dn=kλ2dn = k\lambda

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

2(d+d)n=(k+1)λ2(d+d')n= (k+1)\lambda

d=λ2nd'= \frac{\lambda }{2n}

tanα=db\tan \alpha= \frac{d'}{b}

λ=2nbtanα\lambda = 2nb\tan \alpha

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

λ=21.451034.85105=6.79107m\lambda = 2*1.4*5*10^{-3}*4.85*10^{-5}= 6.79*10^{-7}m


Answer: λ=6.79107m\text{Answer: }\lambda = 6.79*10^{-7}m


 

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