In Young's Double slit experiment, a thin glass mica strip of thickness `(t=8lamda)` is pasted infront of slit `S_(1)`. If the same strip is now shifted in front of other slit `S_(2)`. If the number of fringes which will cross the central point on the screen in `N`. Find `N`? `(dgt gtlamdad` and `n_(mica)=1.5)`

Explanation

To solve the problem, we will analyze the situation step by step.

### Step 1: Understanding the Setup
In Young's Double Slit Experiment, we have two slits, S1 and S2. A thin glass mica strip of thickness \( t = 8\lambda \) is placed in front of slit S1. The refractive index of mica is given as \( n_{mica} = 1.5 \).

### Step 2: Calculate the Optical Path Difference
When light passes through the mica strip, the optical path length increases due to the refractive index. The optical path length for light passing through the mica strip can be calculated as:
\[ \text{Optical Path Length} = n \cdot t \]
For the mica strip:
\[ \text{Optical Path Length} = 1.5 \cdot (8\lambda) = 12\lambda \]
However, since the light in air travels with a refractive index of 1, the effective path difference introduced by the mica strip when placed in front of S1 is:
\[ \Delta x = (n - 1) \cdot t = (1.5 - 1) \cdot (8\lambda) = 0.5 \cdot (8\lambda) = 4\lambda \]

### Step 3: Determine the Number of Fringes
The path difference of \( 4\lambda \) means that at the central point on the screen, the light from S2 will have to travel an additional \( 4\lambda \) to compensate for the path difference created by the mica strip in front of S1.

In Young's experiment, the condition for constructive interference (bright fringes) is given by:
\[ \Delta x = m\lambda \]
where \( m \) is an integer (order of the fringe).

In this case, we have a path difference of \( 4\lambda \), which means that the central maximum (where the path difference is zero) will shift to a new position where the path difference is compensated by the additional \( 4\lambda \).

### Step 4: Counting the Fringes
When the mica strip is moved to S2, the same path difference of \( 4\lambda \) will cause the fringes to shift. The number of fringes that cross the central point can be calculated as:
- The total number of fringes that will cross the central point is equal to twice the number of fringes created by the path difference of \( 4\lambda \).

Since \( 4\lambda \) corresponds to 4 fringes (0, 1, 2, 3, 4), when the strip is moved to S2, these 4 fringes will shift, and thus:
\[ N = 4 \text{ (from S1)} + 4 \text{ (from S2)} = 8 \]

### Final Answer
Thus, the total number of fringes \( N \) that will cross the central point on the screen is:
\[ \boxed{8} \]