Question

In a Young's double slit experiment the intensity at a point is $I$ , where the corresponding path difference is one sixth of the wavelength of light used. If $I_0$ denotes the maximum intensity, the ratio $\frac{I}{I_0}$ is equal to,

• A. $\frac{1}{16}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{3}{16}$

Answer 1

The correct option is C $\frac{3}{4}$

Let the intensity of each slit is $I^{\prime }$

The resultant intensity at a point is,
$I_R=4I^{\prime }{\cos }^2\frac{\varphi }{2}$

Now, $I_{max}=I_0=4I^{\prime }$

and, for the given point, $\varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)=\frac{\pi }{3}(\because \Delta x=\frac{\lambda }{6})$

$\therefore I_R=I=4I^{\prime }{\cos }^2\frac{\pi }{6}=3I^{\prime }=\frac{3I_0}{4}$
Thus,
$\frac{I}{I_0}=\frac{3}{4}$

Hence, $\left(D\right)$ is the correct answer.