Question

In an interference pattern produced by two identical slits, the intensity at the site of the central maximum is $$I$$. The intensity at the same spot when either of the two sites is closed is $$I_0$$. We must have:
A) $$I=I_0$$
B) $$I=2I_0$$
C) $$I=4I_0$$
D) $$I$$ and $$I_0$$ are not related

Answer 1

Hint: In physics, interference is a phenomenon in which two waves superpose to form a resultant wave of greater, lower, or the same amplitude. We have been given the intensities at the site of central maxima for interference pattern from two slits and interference pattern from one slit. The point to be kept in mind while solving this question is that the intensity at the central maxima has the maximum intensity in the entire diffraction pattern.

Formula Used:
$${\mathrm{I}}_{max}={(\sqrt{I_1}+\sqrt{I_2})}^2$$

Complete step by step solution:
Let us consider the intensity of the light from the two slits to be $$I_1$$ and $$I_2$$ respectively. Since we have been told that the slits are identical, we can infer that the intensity of the light through these slits will also be the same. Hence, we can say that $$I_1=I_2=I^{\prime }(let)$$
The intensity of light at central maxima or the maximum intensity is given as $${\mathrm{I}}_{max}={(\sqrt{I_1}+\sqrt{I_2})}^2$$
Substituting the values of the intensities from the two slits, we get,
$$\begin{array}{rl}& {\mathrm{I}}_{max}={(\sqrt{I^{\prime }}+\sqrt{I^{\prime }})}^2\\ & \Rightarrow {\mathrm{I}}_{max}={\left(2\sqrt{I^{\prime }}\right)}^2=4I^{\prime }\end{array}$$
This is the intensity at the central maxima due to interference from both slits. But we have been given this value as $$I$$, hence we can say that $$I=I^{\prime }$$
Now, if one of the slits is closed, the intensity of light through that slit will reduce to zero.
The new maximum intensity will thus be $${\mathrm{I}}_{max}={(\sqrt{I_1}+\sqrt{0})}^2$$
Substituting the value of intensity of light through the slit, we get
$$\begin{array}{rl}& {\mathrm{I}}_{max}={(\sqrt{I^{\prime }}+\sqrt{0})}^2\\ & \Rightarrow {\mathrm{I}}_{max}=I^{\prime }\end{array}$$
But this maximum value has already been given to us as $$I_0$$ .
Hence we can say that $$I_0=I^{\prime }$$.
Now we have a relation of both $$I_0$$ and $$I$$ with $$I^{\prime }$$ as $$I_0=I^{\prime }$$ and $$I=4I^{\prime }$$ respectively.
We can now say that $$I=4I_0$$

Therefore, option (C) is the correct answer.

Note: We have to infer some things from the question on our own. For example, we had been only told that the slits are identical, we inferred on our own that since the slits are identical, they’ll allow equal intensities of light to escape through. If we had not done that, we would have ended up with two equations and three variables and we would never have reached our answer.