Question

In Young’s double-slit experiment, the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

• A. $4:1$
• B. $3:1$
• C. $2:1$
• D. $1:4$

Answer 1

The correct option is A

$4:1$

Step 1: Given data

Here, $w_1\text{and}{w}_2$ are the width of the slits and $D$ is the distance between the screen with two slits and the optical screen.

Step 2: Relationship between the amplitudes

We know from the given information that,

$w_2=3w_1$

From this, we have

$\frac{w_1}{w_2}=\frac{1}{3}$

It is also given that $\mathrm{A}$ is directly proportional to $w$. Then we can write that,

$$\begin{gathered} \therefore \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{w_1}{w_2} \\ \Rightarrow \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}=\frac{1}{3} \\ \Rightarrow 3{\mathrm{A}}_1={\mathrm{A}}_2 \end{gathered}$$

Step 3: Derivation for the required ratio

We know that the formulae maximum and minimum intensities are,

${\mathit{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}{\left(A_1+A_2\right)}^{\mathbf{2}}$, and

${\mathit{I}}_{\mathbf{min}}\mathbf{=}{\mathbf{(}{\mathbf{A}}_{\mathbf{1}}\mathbf{-}{\mathbf{A}}_{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}$

Here, ${\mathrm{A}}_1\text{and}{\mathrm{A}}_2$ are the amplitudes of the light coming from respective slits.

$\begin{array}{rcl}\therefore \frac{I_{max}}{I_{min}}& =& \frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}\\ & =& \frac{{\left(A_1+3A_1\right)}^2}{{\left(A_1-3A_1\right)}^2}\dots \left(\because A_2=3A_1\right)\\ & =& \frac{{\left(4A_1\right)}^2}{{\left(-2A_1\right)}^2}\\ & =& \frac{16{A_1}^2}{4{A_1}^2}\\ & =& 4\end{array}$

Therefore, the ratio is $4:1$.

Hence, the correct option is (A).