1. Use the relation between slit width and amplitude

In Young's double slit experiment, if the two slits have different widths, then the amplitudes of the waves from the slits are proportional to the slit widths.

Given width ratio: $w_1:w_2 = 4:1$ So amplitude ratio is also $a_1:a_2 = 4:1$ Let a_1=4a, a_2=a

2. Intensity at maxima and minima

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

• Maximum intensity: $I_{\max} = (a_1+a_2)^2$
• Minimum intensity: $I_{\min} = (a_1-a_2)^2$

Substitute $a_1=4a$ and $a_2=a$:

$I_{\max}=(4a+a)^2=(5a)^2=25a^2$

$I_{\min}=(4a-a)^2=(3a)^2=9a^2$

3. Find the ratio

$\frac{I_{\max}}{I_{\min}}=\frac{25a^2}{9a^2}=\frac{25}{9}$

Hence, $I_{\max}:I_{\min}=25:9$

4. Check options

• A: $25:9$  Correct
• B: $4:1$  Incorrect
• C: ${\left(\sqrt3+1\right)^4}:16$  Incorrect
• D: $9:1$  Incorrect

Therefore, the correct answer is Option A.