In YDSE, the sources is red ligth of wavelength `7 xx 10^(-7) m`. When a thin glass plate of refractive index 1.5 is put in the path of one of the interfering beams, the central bright fringe shifts by `10^(-3)` m to the position previously occupied by the 5th bright fringe.
If the source is now changed to green light of wavelength `10^(-7)m`, the central fringe shifts to a position initially occupied by the sixth bright fringe due to red ligth. What will be refractive index of glass plate for the second ligth for changed source of ligth?

Explanation

To solve the problem, we will break it down into steps.

### Step 1: Understand the shift caused by the glass plate
When a thin glass plate of refractive index \( \mu \) is placed in the path of one of the beams in Young's Double Slit Experiment (YDSE), the central bright fringe shifts. The formula for the shift \( \Delta x \) is given by:

\[ \Delta x = \frac{(\mu - 1) T D}{d} \]

where:
- \( T \) is the thickness of the glass plate,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

### Step 2: Apply the first case with red light
In the first case, we have red light with a wavelength \( \lambda_r = 7 \times 10^{-7} \) m. The central fringe shifts to the position of the 5th bright fringe, which means:

\[ \Delta x = 5 \frac{D \lambda_r}{d} \]

Setting the two expressions for \( \Delta x \) equal gives:

\[ \frac{(\mu - 1) T D}{d} = 5 \frac{D \lambda_r}{d} \]

Cancelling \( D \) and \( d \) from both sides, we have:

\[ (\mu - 1) T = 5 \lambda_r \]

Substituting \( \mu = 1.5 \) and \( \lambda_r = 7 \times 10^{-7} \):

\[ (1.5 - 1) T = 5 \times 7 \times 10^{-7} \]

\[ 0.5 T = 35 \times 10^{-7} \]

\[ T = \frac{35 \times 10^{-7}}{0.5} = 70 \times 10^{-7} = 7 \times 10^{-6} \text{ m} \]

### Step 3: Apply the second case with green light
Now, we change the source to green light with a wavelength \( \lambda_g = 10^{-7} \) m. The central fringe shifts to the position of the 6th bright fringe of the red light, so:

\[ \Delta x = 6 \frac{D \lambda_r}{d} \]

Using the same formula for the shift:

\[ \Delta x = \frac{(\mu_g - 1) T D}{d} \]

Setting these equal gives:

\[ \frac{(\mu_g - 1) T D}{d} = 6 \frac{D \lambda_r}{d} \]

Cancelling \( D \) and \( d \):

\[ (\mu_g - 1) T = 6 \lambda_r \]

Substituting \( T = 7 \times 10^{-6} \) and \( \lambda_r = 7 \times 10^{-7} \):

\[ (\mu_g - 1) (7 \times 10^{-6}) = 6 \times 7 \times 10^{-7} \]

\[ (\mu_g - 1) (7 \times 10^{-6}) = 42 \times 10^{-7} \]

\[ (\mu_g - 1) (7 \times 10^{-6}) = 4.2 \times 10^{-6} \]

Dividing both sides by \( 7 \times 10^{-6} \):

\[ \mu_g - 1 = \frac{4.2 \times 10^{-6}}{7 \times 10^{-6}} = 0.6 \]

Thus,

\[ \mu_g = 1.6 \]

### Final Answer
The refractive index of the glass plate for the green light is \( \mu_g = 1.6 \).

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