Question Details

The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum of the minimum intensity in the interference pattern is :

Options

9:1

16:1

1:1

4:1

Show Answer

Correct Answer :

Option A

9:1

Solution :

The correct answer is Option 1: 9:1.

In Young's Double Slit Experiment (YDSE), the intensity of light from each slit is directly proportional to the width of that slit. This is because a wider slit allows more light through, increasing the amplitude of the wave, and intensity is proportional to the square of the amplitude.

Let the width of the narrower slit be w. Then the width of the wider slit is 4w.

Step 1: Find the ratio of intensities from the two slits.

Since intensity ∝ slit width:

$\frac{I_{1}}{I_{2}} = \frac{4 w}{w} = \frac{4}{1}$

Step 2: Find the ratio of amplitudes.

Since intensity ∝ (amplitude)², we have amplitude ∝ √intensity:

$\frac{a_{1}}{a_{2}} = \sqrt{\frac{I_{1}}{I_{2}}} = \sqrt{\frac{4}{1}} = \frac{2}{1}$

So let a1 = 2 (amplitude from the wider slit) and a2 = 1 (amplitude from the narrower slit).

Step 3: Calculate maximum intensity.

Maximum intensity occurs when the two waves interfere constructively (phase difference = 0):

$I_{\max} = (a_{1} + a_{2})^{2} = (2 + 1)^{2} = 9$

Step 4: Calculate minimum intensity.

Minimum intensity occurs when the two waves interfere destructively (phase difference = π):

$I_{\min} = (a_{1} - a_{2})^{2} = (2 - 1)^{2} = 1$

Step 5: Find the ratio.

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

Therefore, the ratio of maximum to minimum intensity is 9 : 1.

Key Insight: Notice that if both slits had equal widths (a₁ = a₂), destructive interference would produce zero minimum intensity (I_min = 0). The fact that the slits have unequal widths means their amplitudes are unequal (2 ≠ 1), so destructive interference is only partial, giving a non-zero minimum intensity of 1 unit. This is why the ratio 9:1 is obtained rather than ∞.