In young'sdouble slit exp. if glass slab is placed in front of one slit. wyh pathdiffrence becomes (u-1)t.

Question

where u is the refrctive index of glass and t is the thickness

Answer 1

Before the glass slab is placed in front of one slit:

X = λL/d, where λ is the wavelength of the light, L is the distance from the slit to the background screen, d is the separation of the slit and X is the distance from the central maxima to the first band of the interference pattern.

After the glass slab is placed in front of a slit, the wavelength λ is changed due to the refraction index of the glass. Hence the path difference of the light is given by:

ΔX = X(a) - X(g) = L/d [λ(a)-λ(g)], where X(g) is the distance from the central maxima to the first band of the interference pattern with the glass slab, and X(a) is without the glass slab. λ(g) is the changed wavelength of light due to the glass slab, and λ(a) is the original wavelength without the glass slab.

But we also know that u, the index of fraction of the glass is given by:

u = c/v, where c is the speed of light in vacuum and v is the average velocity of light in the glass slab.

u = c/v = fλ(a)/fλ(g) = λ(a)/λ(g) or λ(a)=uλ(g)

Therefore, ΔX = L/d [λ(a)-λ(g)] = (L/d) [uλ(g)-λ(g)]

= (L/d) λ(g)(u-1), but (L/d)λ(g) is just t, the thickness of the glass

Hence, ΔX = (u-1)t

Answer 2

This is becuase the speed of light in all the material media is not he same. To insert the correction factor in the path diference which v derive it comes uout (u-1)t as the distance travelled depend on the velocity of light and the velocity of light is the factor on which refractive index depends. Remember the Snells law.

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