Question 53: Wave Optics

Question

The maximum intensity in a Young's double slit experiment is $I_0$. Distance between the slits $(d)$ is $5 \lambda$, where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D=10 d$ is $\_\_\_\_$ .

Accepted Answer

In YoungÃ¢ÂÂs double slit experiment, intensity at any point on the screen is
$$ I = I_0 \cos^2 \left(\frac{\phi}{2}\right) $$
where $I_0$ is the maximum intensity and $\phi$ is the phase difference.
Now we need the intensity at the point on the screen exactly opposite to one of the slits.
Let the slits be $S_1$ and $S_2$, separated by distance $d$.
At the point directly opposite to slit $S_1$:
distance from $S_1$ to that point $= D$
distance from $S_2$ to that point $= \sqrt{D^2 + d^2}$
So, path difference is
$$ \Delta = \sqrt{D^2+d^2}-D $$
Given:
$$ D = 10d $$
Therefore,
$$ \Delta = \sqrt{(10d)^2+d^2}-10d = \sqrt{101d^2}-10d = d(\sqrt{101}-10) $$
Since $d = 5\lambda$,
$$ \Delta = 5\lambda(\sqrt{101}-10) $$
Now,
$$ \sqrt{101} \approx 10.05 $$
So,
$$ \Delta \approx 5\lambda(10.05-10)=5\lambda(0.05)=0.25\lambda=\frac{\lambda}{4} $$
Hence phase difference is
$$ \phi = \frac{2\pi}{\lambda}\Delta = \frac{2\pi}{\lambda}\cdot \frac{\lambda}{4} = \frac{\pi}{2} $$
Now intensity becomes
$$ I = I_0 \cos^2\left(\frac{\phi}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} $$
So the required intensity is
$$ \boxed{\frac{I_0}{2}} $$
Therefore, the correct option is
B
.

Answer

$$ \frac{I_0}{4} $$

Answer

$$ \frac{I_o}{2} $$

Answer

$I_{\mathrm{o}}$

Answer

$$ \frac{3 I_0}{4} $$

Wave Optics

The maximum intensity in a Young's double slit experiment is I_0. Distance between the slits (d) is 5 \lambda, where \lambda is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10 d is \_\_\_\_ .

Choose the correct answer: