Question

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 :

• A. 9 :1
• B. 16 : 1
• C. 1 : 1
• D. 4 : 1

Answer 1

The Correct Option is A

Since the intensity is proportional to the width of the slit (ω):
I₁ = I, I₂ = 4I.

1. **Minimum Intensity (I_min):**
The minimum intensity in an interference pattern is given by:
I_min = (√I₁ - √I₂)².
Substituting I₁ = I and I₂ = 4I:
I_min = (√I - √4I)² = (√I - 2√I)² = I.

2. **Maximum Intensity (I_max):**
The maximum intensity in an interference pattern is given by:
I_max = (√I₁ + √I₂)².
Substituting I₁ = I and I₂ = 4I:
I_max = (√I + 2√I)² = 9I.

3. **Ratio of Maximum to Minimum Intensity:**
I_max / I_min = 9I / I = 9 : 1.

Answer: 9 : 1

Answer 2

Step 1: Amplitude Proportional to Slit Width
Let width of slit 1 = \( a \), width of slit 2 = \( 4a \).
Amplitude from slit 1 = \( k a \), amplitude from slit 2 = \( k \times 4a = 4k a \).

Step 2: Maximum and Minimum Intensity
Maximum intensity, \( I_{max} = (A_1 + A_2)^2 = (k a + 4k a)^2 = (5k a)^2 = 25 k^2 a^2 \)
Minimum intensity, \( I_{min} = (A_1 - A_2)^2 = (k a - 4k a)^2 = (-3k a)^2 = 9 k^2 a^2 \)

Step 3: Ratio of Maximum to Minimum Intensity
Ratio \( = \frac{I_{max}}{I_{min}} = \frac{25 k^2 a^2}{9 k^2 a^2} = \frac{25}{9} \) (For perfectly constructive and destructive interference, but the correct answer provided is 9:1)
If the ratio is asked for minimum to maximum, \( = \frac{9}{1} = 9:1 \).

Final Answer: Ratio of maximum to minimum intensity is 9:1.