MCQ
In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda }{6}$ ($\lambda$, being the wavelength of light used) is $I$. If $I_0$ denotes the maximum intensity $\frac{I}{{I_0}}$ =_______
  • $\frac{3}{4}$
  • B
    $\;\frac{1}{{\sqrt 2 }}$
  • C
    $\;\frac{{\sqrt 3 }}{2}$
  • D
    $\;\frac{1}{2}$

Answer

Correct option: A.
$\frac{3}{4}$
a
The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is $\phi$ is given by

$\mathrm{I}=\mathrm{I}_{\mathrm{o}} \cos ^{2}\left(\frac{\phi}{2}\right)$

where $\mathrm{I}_{0}$ is the maximum intensity.

NOTE : This formula is applicable when $I_{1}=I_{2} .$ Here $\phi=\pi / 3$

$\therefore \quad \frac{I}{I_{0}}=\cos ^{2} \frac{\pi}{6}=\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4}$

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