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46.The 7th bright fringe in YDSE using a light of wavelength 550 nm shifts to the cen- tral maxima after covering the two slits with two sheets of difference refractive indices n1 and n2 but having same thickness 6 µm. The value of —n1 − n2— is (integer value?
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46.The 7th bright fringe in YDSE using a light of wavelength 550 nm sh...
To solve this problem, we need to apply the concept of interference in Young's double-slit experiment. Let's break down the problem into smaller steps:

Step 1: Find the fringe width (β)
The fringe width (β) can be calculated using the formula:
β = λD / d
Where:
- λ is the wavelength of light (550 nm)
- D is the distance between the screen and the double slits
- d is the distance between the two slits

Step 2: Find the distance (x) covered by the 7th bright fringe
The distance (x) covered by the 7th bright fringe can be calculated using the formula:
x = 7β

Step 3: Find the shift in the position of the 7th bright fringe after covering the slits
The shift in the position of the 7th bright fringe can be calculated using the formula:
Δx = (n1 - n2) * t
Where:
- n1 is the refractive index of the first sheet
- n2 is the refractive index of the second sheet
- t is the thickness of the sheets (6 µm)

Step 4: Equate the shift in the position of the 7th bright fringe to the distance covered by the fringe
Δx = x

Step 5: Solve for |n1 - n2|
|n1 - n2| = x / t

Using the values calculated in the previous steps, we can substitute them into the equation to find the value of |n1 - n2|.
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46.The 7th bright fringe in YDSE using a light of wavelength 550 nm shifts to the cen- tral maxima after covering the two slits with two sheets of difference refractive indices n1 and n2 but having same thickness 6 µm. The value of —n1 − n2— is (integer value?
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