1. Relate slit width to amplitude

In Young’s double slit experiment, the amplitude contributed by a slit is proportional to its width.

Let the wider slit have width $2w$ and the narrower slit have width $w$. Then the amplitudes are in the ratio $a_1:a_2 = 2:1.$

So, let $a_1=2a, \qquad a_2=a.$

2. Expression for maximum and minimum intensity

For two coherent sources with amplitudes $a_1$ and $a_2$:

• Maximum intensity occurs when the waves interfere constructively: $I_{\max}=(a_1+a_2)^2$

• Minimum intensity occurs when the waves interfere destructively: $I_{\min}=(a_1-a_2)^2$

Substitute $a_1=2a$ and $a_2=a$:

$I_{\max}=(2a+a)^2=(3a)^2=9a^2$

$I_{\min}=(2a-a)^2=a^2$

3. Find the ratio

$\frac{I_{\max}}{I_{\min}}=\frac{9a^2}{a^2}=9$

Hence, $I_{\max}:I_{\min}=9:1.$

4. Check options

• A: $3:1$ ❌
• B: $(2\sqrt{2}+1):(2\sqrt{2}-1)$ ❌
• C: $9:1$ ✅
• D: $(3+2\sqrt{2}):(3-2\sqrt{2})$ ❌

So the correct option is C.

5. Comparison with stored answer

The stored correct answer is D, but the derived answer is C.

The likely reason: intensity is proportional to the square of amplitude, while amplitude is proportional to slit width. Since one slit width is half the other, amplitudes are in ratio $1:2$, not intensities. Therefore using $\frac{I_{\max}}{I_{\min}}=\frac{(2+1)^2}{(2-1)^2}=9$ is the correct approach.