Question

In Youngs double slit experiment, the central fringe of interference pattern produced by light of wavelength $6000\,\overset{\text{o}}{\mathop{\text{A}}}\,$ shifts to the position of the 5th bright fringe on introducing a thin glass plate of refractive index of 1.50 in front of one of the slits. The thickness of the plate would be:

• A. $\text{6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm}$
• B. $\text{4 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm}$
• C. $\text{5 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm}$
• D. $\text{6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{cm}$

Answer 1

Answer: (A)

$\text{6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm}$

Answer 2

The shift in the fringe pattern – 4λD/d

Here, λ = 6000x10-10m

Now, shift due to glass plate= (μ-1)tD/d

Now, equating both the equations–

5λD/d = (μ-1)tD/d

So, t= 5λ/(μ-1)

t= 5x6000x10-10/ (1.5-1) = 6x10-6m = 6x10-4cm
The light fringe is referred to as the core fringe for N=0. Along the core fringe, the higher-order fringes are asymmetrically placed. The nth fringe indicates where the brilliant fringe is located.

Y(brightness) = (nπ\d)D (n=0, n=+1 -1, +2 -2….)

Fringe’s with minimum intensity: Y(dark) = (2n-1)πD/2d (n=0, n=+1 -1, +2 -2….)

Fringe width is the distance between two adjacent bright and adjacent dark fringes.

Let's consider the two fringes between position n and n+1

So, fringe width = ((n+1 π\d ) D - (nπ\d)D = πD/ d

The distance of the nth bright fringe from the center is

xn = nλD/d

The distance of the nth dark fringe from the center is:

xn = (2n+1)λD/2d