Q124 / 150 · MHT CET 2026 (16 April Shift 2) · Physics

In Young's double slit experiment, two slits are illuminated with a light of wavelength $\lambda$. The line joining $A_1P$ is perpendicular to $A_1A_2$ as shown in figure. If the first minimum is detected at P, the value of slits separation 'a' will be
(D = distance between source and screen)
Figure

  • A
    $\lambda\text{D}$
  • Correct answer
    B
    $\sqrt{\lambda\text{D}}$
  • C
    $\sqrt{\dfrac{\lambda}{\text{D}}}$
  • D
    $\sqrt{\dfrac{\text{D}}{\lambda}}$

Solution

From the given figure, the path difference at point P is given by:

$\Delta x = A_2P - A_1P$

Using Pythagoras theorem in right-angled triangle $A_1A_2P$, we have $A_1P = D$ and $A_2P = \sqrt{D^2 + a^2}$.

So, $\Delta x = \sqrt{D^2 + a^2} - D$

For the first minimum to be detected at P, the path difference must be $\dfrac{\lambda}{2}$.

$\sqrt{D^2 + a^2} - D = \dfrac{\lambda}{2}$

$\sqrt{D^2 + a^2} = D + \dfrac{\lambda}{2}$

Squaring both sides, we get:

$D^2 + a^2 = D^2 + D\lambda + \dfrac{\lambda^2}{4}$

$a^2 = D\lambda + \dfrac{\lambda^2}{4}$

Since the wavelength of light $\lambda$ is very small compared to the distance $D$ ($\lambda \ll D$), the term $\dfrac{\lambda^2}{4}$ is negligibly small and can be ignored.

$a^2 \approx D\lambda$

$a = \sqrt{\lambda D}$

Answer: $\sqrt{\lambda\text{D}}$

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

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