In a Young's double slit interference pattern at a point , we observe the 10th bright fringe (order maxima) for wavelength 7000 Å . What order maxima will be visible if the source of light is replaced by light of wavelength 5000 Å?

Answer

To solve the problem step by step, we will use the relationship between the order of maxima, the wavelength of light, and the position of the fringes in a Young's double slit interference pattern.

### Step 1: Understand the relationship between order of maxima and wavelength
In a Young's double slit experiment, the condition for bright fringes (maxima) is given by the formula:
\[ N \lambda = d \sin \theta \]
where:
- \( N \) is the order of the maxima,
- \( \lambda \) is the wavelength of light,
- \( d \) is the distance between the slits,
- \( \theta \) is the angle of the fringe from the central maximum.

For small angles, we can simplify this to:
\[ N = \frac{y}{\lambda} \]
where \( y \) is the distance from the central maximum to the \( N^{th} \) bright fringe.

### Step 2: Set up the equation for the first scenario
We know that at a point, the 10th bright fringe is observed for a wavelength of \( 7000 \) Å. Thus, we can write:
\[ N_1 \lambda_1 = N_2 \lambda_2 \]
where:
- \( N_1 = 10 \) (the order of maxima for the first wavelength),
- \( \lambda_1 = 7000 \) Å (the first wavelength),
- \( N_2 \) is the order of maxima we need to find,
- \( \lambda_2 = 5000 \) Å (the second wavelength).

### Step 3: Rearranging the formula
From the relationship, we can rearrange it to find \( N_2 \):
\[ N_2 = \frac{N_1 \lambda_1}{\lambda_2} \]

### Step 4: Substitute the values
Now we can substitute the known values into the equation:
\[ N_2 = \frac{10 \times 7000 \, \text{Å}}{5000 \, \text{Å}} \]

### Step 5: Calculate \( N_2 \)
Perform the calculation:
\[ N_2 = \frac{70000}{5000} = 14 \]

### Conclusion
Thus, the order of maxima that will be visible when the source of light is replaced by light of wavelength \( 5000 \) Å is \( 14 \).