Answer :

Given that, d = 0.50 m m u00a0= u00a00.5 u00a0u00d7 u00a010 u2212 3 m u00a0Diameter hole is illuminated by a light of wavesize u00a0510 n m u00a0= u00a0510 u00a0u00d7 10 ^u2212 9 m The angle subtended by the maxima is provided by u00a0sin u03b8 u00a0= u00a0m u03bb/ d u00a0Wbelow m = 1,2,3,4..... for the positions of various maxima The angle subtfinished by the first main 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 => u00a0 1.020 u00d7 10 u2212 3 u00a0 Now from the diagram, we can see u00a0Y u00a0= u00a0L u00a0tan u03b8 u00a0For little angle u00a0u03b8 u00a0we have u00a0sin u03b8 u00a0u223c u00a0tan u03b8 u00a0 So, u00a0Y u00a0= u00a0L u00a0sin u03b8 => u00a02.2 u00a0u00d7 u00a01.020 u00a0u00d7 u00a010^ u2212 3 u00a0=> u00a02.244 u00a0u00d7 u00a010^ u2212 3 m u00a0 The full width of the main maximum in the diffractivity pattern on the display u00a0= > 2 Y u00a0= u00a02 u00a0u00d7 u00a02.93 u00d7 10^ u2212 3 m => 4.488 u00a0u00d7 u00a010^ u2212 3 m => 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 hole is illuminated by a light of wavelength

510 n m = 510 ×

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The angle subtended by the maxima is given by

sin θ = m λ/ d

Wbelow m = 1,2,3,4..... for the positions of various maxima

The angle subtended by the first main maxima is provided by

sin θ = 1 × λ /d

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

=> 1.020 × 10 − 3

Now from the diagram, we deserve to watch

Y = L tan θ

For small 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 full width of the main maximum in the diffraction pattern on the screen

= > 2 Y = 2 × 2.93 × 10^ − 3 m

=> 4.488 × 10^ − 3 m

=> 5.86 mm


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