Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of t

Question

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:

Accepted Answer

Given: Width of one slit is half of the other in Young's double slit experiment.

Find: The ratio of maximum intensity to minimum intensity.

Intensity due to a slit is directly proportional to its width. Let the slit widths be $w$ and $2w$ . Then the corresponding intensities may be taken as $I_1 = I_0$ and $I_2 = 2I_0$ .

For interference of two coherent sources,

I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2

and

I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2

Substituting $I_1 = I_0$ and $I_2 = 2I_0$ ,

I_{\max} = (\sqrt{I_0} + \sqrt{2I_0})^2 = I_0(1+\sqrt{2})^2 = I_0(3+2\sqrt{2})

I_{\min} = (\sqrt{I_0} - \sqrt{2I_0})^2 = I_0(1-\sqrt{2})^2 = I_0(3-2\sqrt{2})

Therefore,

\frac{I_{\max}}{I_{\min}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}

So, the ratio of maximum to minimum intensity is $(3 + 2\sqrt{2}) : (3 - 2\sqrt{2})$ . The correct option is B.

Given: One slit has half the width of the other.

Find: $I_{\max} : I_{\min}$ .

Let the wider slit have width $w$ and the narrower slit have width $\frac{w}{2}$ . Since intensity is proportional to slit width,

I_1 : I_2 = w : \frac{w}{2} = 2 : 1

Hence,

\sqrt{I_1} : \sqrt{I_2} = \sqrt{2} : 1

Now use the standard expressions,

I_{\max} = (\sqrt{I_1}+\sqrt{I_2})^2

I_{\min} = (\sqrt{I_1}-\sqrt{I_2})^2

Taking the ratio form directly,

I_{\max} : I_{\min} = (\sqrt{2}+1)^2 : (\sqrt{2}-1)^2 = (3+2\sqrt{2}) : (3-2\sqrt{2})

Therefore, the correct option is B.

Answer

Step-by-step solution

Common mistakes

Practice more Interference (Young's Experiment) questions

Related questions