Q70 / 75 · JEE Main 2026 (06 April Shift 2) · Physics

In a Young double slit experiment, the wavelength of incident light is $6000$ Å, the separation between slits $S_1$ and $S_2$ is $5$ cm and the distance between slits plane and screen is $50$ cm, as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is __________ Å.
Figure

  • A
    $4000$
  • B
    $3000$
  • Correct answer
    C
    $2000$
  • D
    $1000$

Solution

Let the intensity due to each individual slit be $I_0$.
The resultant intensity $I_R$ at a point where the phase difference is $\phi$ is given by:

$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$

Given that $I_1 = I_2 = I_0$ and the resultant intensity $I_R = I_0$, we can substitute these values into the equation:

$I_0 = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos \phi$

$I_0 = 2I_0 + 2I_0 \cos \phi$

$I_0 = 2I_0 (1 + \cos \phi)$

$1 + \cos \phi = \dfrac{1}{2}$

$\cos \phi = -\dfrac{1}{2}$

The minimum phase difference satisfying this condition is:
$\phi = \dfrac{2\pi}{3}$

The relationship between phase difference $\phi$ and path difference $\Delta x$ is:
$\phi = \dfrac{2\pi}{\lambda} \Delta x$

Substituting the value of $\phi$:
$\dfrac{2\pi}{3} = \dfrac{2\pi}{\lambda} \Delta x$

$\Delta x = \dfrac{\lambda}{3}$

Given the wavelength $\lambda = 6000 \text{ \AA}$:
$\Delta x = \dfrac{6000}{3} = 2000 \text{ \AA}$

Answer: $2000$

Hence, the correct answer is option: 'C'.

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