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