In a Young's double slits experiment, the ratio of amplitude of light coming from slits is 2:1. The ratio of the maximum to minimum intensity in the interference pattern is

Question

Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same.
Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

Answer 1

The problem given is about the Young's double slit experiment, where we need to find the ratio of maximum to minimum intensity in the interference pattern given the amplitude ratio of light from the slits.

The key concept here involves the relationship between the amplitudes of the light waves and the resulting intensities. In Young's double slit experiment, the intensity of the resultant light is related to the amplitude of the light waves from the two slits.

Let's denote the amplitudes of light from the two slits as \( A_1 \) and \( A_2 \). According to the problem, the ratio of the amplitudes is given by:

A_1 : A_2 = 2 : 1

The intensity of light is proportional to the square of the amplitude, hence:

I_1 \propto A_1^2 and I_2 \propto A_2^2

The resultant intensity \( I \) due to interference pattern is given by:

I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi

where \( \phi \) is the phase difference between the two waves. The maximum intensity occurs when \( \cos\phi = 1\):

I_{max} = (A_1 + A_2)^2

The minimum intensity occurs when \( \cos\phi = -1\):

I_{min} = (A_1 - A_2)^2

Substituting the amplitude ratio into these formulas, we find:

I_{max} = (2A + 1A)^2 = (3A)^2 = 9A^2
I_{min} = (2A - 1A)^2 = (1A)^2 = A^2

Therefore, the ratio of maximum to minimum intensity is:

\frac{I_{max}}{I_{min}} = \frac{9A^2}{A^2} = 9:1

Thus, the correct answer is 9:1.

Answer 2