1. In Young’s double slit experiment, if the amplitudes of the two waves are $a_1$ and $a_2$, then:

$I_{\max} = (a_1+a_2)^2$ $I_{\min} = (a_1-a_2)^2$

2. Given amplitude ratio:

$a_1:a_2 = 2:1$

Let

$a_1=2a, \quad a_2=a$

3. Maximum intensity:

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

4. Minimum intensity:

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

5. Therefore, the ratio is:

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

So,

$I_{\max}:I_{\min} = 9:1$

6. Comparing with the options:

• A: $25:9$ ❌
• B: $9:1$ ✅
• C: $9:4$ ❌
• D: $2:1$ ❌

Hence, the correct answer is Option B.