Question

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.21 m away producing an interference pattern. A piece of glass with index of refraction n = 1.79 is placed at one slit. Placing the piece of glass in front of the slit causes the maxima to shift 0.37δx, where δx is the distance between adjacent maxima. What is the thickness of the glass in μm?

Answer 1

Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$