Relate slit width to amplitude

Let the smaller slit have width $w$.

Then the larger slit has width $3w$.

Since amplitude is proportional to slit width, $a_1 : a_2 = 1 : 3$

So we can take $a_1 = a, \qquad a_2 = 3a$

Use formulas for maximum and minimum intensity

In Young's double slit experiment, if the amplitudes from the two slits are $a_1$ and $a_2$, then:

• Maximum resultant amplitude: $A_{\max} = a_1 + a_2$
• Minimum resultant amplitude: $A_{\min} = |a_2 - a_1|$

Since intensity is proportional to square of amplitude, $I_{\max} \propto (a_1+a_2)^2$ $I_{\min} \propto (a_2-a_1)^2$

Substitute the amplitudes

$I_{\max} : I_{\min} = (a+3a)^2 : (3a-a)^2$

$= (4a)^2 : (2a)^2$

$= 16a^2 : 4a^2$

$= 4 : 1$

Match with the options

The correct option is: $\boxed{\text{A: } 4:1}$

Comparison with stored answer

Stored correct answer is A.

Our derived answer is also A. So the stored answer is correct.