Answer :\n

Given that,\nd = 0.50 m m \u00a0= \u00a00.5 \u00a0\u00d7 \u00a010 \u2212 3 m \u00a0Diameter hole is illuminated through a light of wavelength \u00a0510 n m \u00a0= \u00a0510 \u00a0\u00d7 10 ^\u2212 9 m <\/tex>\nThe angle subtended by the maxima is offered by \u00a0sin \u03b8 \u00a0= \u00a0m \u03bb\/ d \u00a0Where m = 1,2,3,4..... Because that the location of different maxima\nThe edge subtended by the first central maxima is provided by \u00a0sin \u03b8 \u00a0= \u00a01 \u00d7 \u03bb \/d \u00a0=> \u00a0(1 \u00d7 510 \u00d7 10 ^\u22129) \/ \u00a0(0.50 \u00d7 10 ^\u2212 3) \u00a0\n=> \u00a0 1.020 \u00d7 10 \u2212 3 \u00a0\nNow indigenous the diagram, we deserve to see \u00a0Y \u00a0= \u00a0L \u00a0tan \u03b8 \u00a0For small angle \u00a0\u03b8 \u00a0we have actually \u00a0sin \u03b8 \u00a0\u223c \u00a0tan \u03b8 \u00a0\nSo, \u00a0Y \u00a0= \u00a0L \u00a0sin \u03b8\n=> \u00a02.2 \u00a0\u00d7 \u00a01.020 \u00a0\u00d7 \u00a010^ \u2212 3 \u00a0=> \u00a02.244 \u00a0\u00d7 \u00a010^ \u2212 3 m \u00a0\nThe total width that the central maximum in the diffraction pattern on the display screen \u00a0= > 2 Y \u00a0= \u00a02 \u00a0\u00d7 \u00a02.93 \u00d7 10^ \u2212 3 m\n=> 4.488 \u00a0\u00d7 \u00a010^ \u2212 3 m\n=> \u00a05.86 \u00a0mm"}>" data-test="answer-box-list">


You are watching: What is the width of the central maximum on a screen 2.2 m behind the slit?

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Answer :

Given that,

d = 0.50 m m = 0.5 × 10 − 3 m

Diameter feet is illuminated through a light of wavelength

510 n m = 510 ×

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The angle subtended by the maxima is offered by

sin θ = m λ/ d

Where m = 1,2,3,4..... For the positions of various maxima

The edge subtended by the first main maxima is offered by

sin θ = 1 × λ /d

=> (1 × 510 × 10 ^−9) / (0.50 × 10 ^− 3)

=> 1.020 × 10 − 3

Now indigenous the diagram, we can see

Y = together tan θ

For tiny angle θ we have sin θ ∼ tan θ

So,

Y = l sin θ

=> 2.2 × 1.020 × 10^ − 3

=> 2.244 × 10^ − 3 m

The total width that the central maximum in the diffraction sample on the display

= > 2 Y = 2 × 2.93 × 10^ − 3 m

=> 4.488 × 10^ − 3 m

=> 5.86 mm


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a diffraction grating used at normal incidence gives a line(540 nm) in a specific order superposed ~ above the violet heat (405nm) that the next greater order.…how plenty of lines per cm room there in the lattice if the edge of diffraction is 30°?