In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is `lambda` is `K,(lambda` being the wave length of light used). The intensity at a point where the path difference is `(lambda)/(4)`, will be `(K)/(n)`. Find `n`.
'(##CEN_KSR_PHY_JEE_C27_E01_007_Q01##)'

Answer

To solve the problem, we will use the relationship between path difference, phase difference, and intensity in the context of Young's double-slit experiment.

### Step 1: Understand the relationship between path difference and phase difference
The phase difference \(\phi\) is related to the path difference \(d\) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \cdot d \]
where \(\lambda\) is the wavelength of the light.

### Step 2: Calculate the phase difference for the first case
For the first case, where the path difference \(d = \lambda\):
\[ \phi_1 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]

### Step 3: Calculate the intensity for the first case
The intensity \(I\) at this point is given as \(K\). In terms of the intensity of the individual slits \(I_0\), we can express the total intensity as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
Assuming \(I_1 = I_2 = I_0\), we have:
\[ I = 2I_0 + 2\sqrt{I_0 I_0} \cos(2\pi) = 2I_0 + 2I_0 = 4I_0 \]
Thus, we can equate:
\[ K = 4I_0 \quad \Rightarrow \quad I_0 = \frac{K}{4} \]

### Step 4: Calculate the phase difference for the second case
For the second case, where the path difference \(d = \frac{\lambda}{4}\):
\[ \phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

### Step 5: Calculate the intensity for the second case
Using the same formula for intensity:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi_2) \]
Substituting \(I_1 = I_2 = I_0\):
\[ I = 2I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{\pi}{2}\right) = 2I_0 + 0 = 2I_0 \]
Now substituting \(I_0 = \frac{K}{4}\):
\[ I = 2 \cdot \frac{K}{4} = \frac{K}{2} \]

### Step 6: Relate the intensity to the given condition
According to the problem, the intensity at this point is given as:
\[ I = \frac{K}{n} \]
Setting the two expressions for intensity equal:
\[ \frac{K}{2} = \frac{K}{n} \]

### Step 7: Solve for \(n\)
To find \(n\), we can cross-multiply:
\[ K \cdot n = 2K \quad \Rightarrow \quad n = 2 \]

### Final Answer
Thus, the value of \(n\) is:
\[ \boxed{2} \]