Light with a wavelength 446 nm passes through a double-slit system that has a sl

Question

Light with a wavelength 446 nm passes through a double-slit system that has a slit separation d = 0.800 mm. Determine how far away a screen must be placed in order that a dark fringe appear directly opposite both slits, with just one bright fringe between them. cm A lens made of glass (n_g = 1.52) is coated with a thin film of SiO (n_s = 1.45) of thickness t. Visible light is incident normally on the coated lens as in the figure below. (a) For what minimum value of t will the reflected light of wavelength 535 nm (in air) be missing? nm (b) Are there other values of t that will minimize the reflected light at this wavelength?

Explanation / Answer

1)

Using Young's double slit equation we know that L = 2*d*y/lamda = (2*0.8*10^-3*0.4*10^-3)/(446*10^-9)

where d is the distnace between slits = 0.8 mm, y is the distance from central bright and next dark fringe = 0.4 mm

L = 1.435 m = 143.5 cm

2)

a) n2=1.52; n1=1.45; lambda0=535nm

destructive interference condition 2d=(m+0.5)lambda1=(m+0.5)lambda0/n1 where m=0,1,2,3...

For m=0, we get d=(m+0.5)lambda0/2n1=lambda0/4n1=535nm/4*1.45=92.24nm

Minimum value of t =92.24nm

b) yes, all the other values than m can have besides 0, this is m=1,2,3...

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