Two slits spaced 00720 mathrmmm apart are 0800 mathrmm from...

Two slits spaced $0.0720 \mathrm{~mm}$ apart are $0.800 \mathrm{~m}$ from a screen. Coherent light of wavelength $\lambda$ passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is $3.00 \mathrm{~mm} .$ If the intensity at the peak of the central maximum is $0.0600 \mathrm{~W} / \mathrm{m}^{2},$ what is the intensity at points on the screen that are (a) $2.00 \mathrm{~mm}$ and (b) $1.50 \mathrm{~mm}$ from the center of the central maximum?

Transcript

00:01 Here in this problem, we have to find out the intensity at different points.

00:08 So first let's see that what is intensity? so the intensity is given as i equal i0, course square by d divided by lambda.

00:41 L multiply by y now here i note is the maximum intensity lambda is the wavelength d is the spacing between the slits capital l is the distance between screen and slits and y is the distance between the first minimum and the central maximum now the distance between the first minimum and the central maximum is given as y equal to lambda l divided by d multiplied by m plus one divided by now as we know that the first minimum occurs at m equals 0 thus y is equal to lambda l divided by 2d now let's rewrite this for lambda so we get lambda is equal to 2 y d divided by capital l now here we substitute 3 millimeter for y 0 .0720 millimeter for d and 0 .800 meter for d for for l so we get lambda is equal to 2 multiply by 3 millimeter now we do one thing just we convert this into meter also so let's multiply it by 10 to the power minus 3 multiply by 0 .0720 millimeter we also convert this into meter so 10 to the power minus 3 meter divided by 0 .80 meter.

03:30 Now from above we get lambda is equal to 5 .4 multiply by 10 to the power minus 7 meter.

03:42 Now comes to part a.

03:46 So here we have to find out the intensity at y equal millimeter.

03:58 At y equal 2 millimeter the intensity at the point on the screen 2 millimeter from center of the central maximum is given by i equal i0 course square y d multiply by lambda l multiply by y now here we substitute 2 millimeter for y 5 .4 multiply by 10 to the power minus 7 meter for lambda, 0 .0720 millimeter for d and 0 .8080 meter for l and 0 .0600 watt per meter square for i node.

04:52 Therefore we get i equal to 0 .06 .2 watt per meter square for i note.

05:00 X00 multiply by course square 3 .14 multiply by 0 .0720 multiply by 10 to the power minus 3.

05:21 Multiply by 2 multiply by 10 to the power minus 3...

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