Question

In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit- width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

पर्याय

• 4 : 1

• 2 : 1

• 3 : 1

• 1 : 4

Answer 1

4 : 1

Explanation:

Given,

Let the width of the slit be ω

So, ω₁ = 3ω₂

Since, A ∝ ω

Ratio of maximum to minimum intensity

`I_"max"/I_"min"` = ?

I = `"A"_1^2 + "A"_2^2 + 2"A"_1 +"A"_2` cos Φ

For maximum intensity cos Φ = 1

For minimum intensity cos Φ = - 1

So, I_max = (A₁ + A₂)² 

I_min = (A₁ - A₂)² 

Since, A₁ = A and A₂ = 3A

So, `"A"_1/"A"_2 = omega_1/omega_2`

Hence, `"I"_"max"/"I"_"min" = (("A"+3"A")^2)/(("A"-3"A")^2)`

⇒  `"I"_"max"/"I"_"min" = (4"A")^2/(-2"A")^2`

⇒  `"I"_"max"/"I"_"min" = 16/4 = 4/1` 

Hence `"I"_"max"/"I"_"min" = 4/1`