## A 0.40 mm diameter hole is illuminated by a light of wavelength 510 n m . What is the width of the central maximum in a screen 2.3 m behind the slit?

Question

### Question:

A 0.40 mm diameter hole is illuminated by a light of wavelength 510 n m . What is the width of the central maximum in a screen 2.3 m behind the slit?

Given that,
d = 0.50 m = 0.5 × 10 − 3 m
Breadth opening is lit up by a light of wavelength
510 n m = 510 ×
The point subtended by the maxima is given by
sin θ = m λ/d
Where m = 1,2,3,4….. for the places of various maxima
The edge subtended by the primary focal maxima is given by
sin θ = 1 × λ/d
=> (1 × 510 × 10 ^−9)/(0.50 × 10 ^− 3)
=> 1.020 × 10 − 3
Presently from the chart, we can see
Y = L tan θ
For little edge θ we have sin θ ∼ tan θ
Along these lines,
Y = L sin θ
=> 2.2 × 1.020 × 10^ − 3
=> 2.244 × 10^ − 3 m
The absolute width of the focal greatest in the diffraction design on the screen
= > 2 Y = 2 × 2.93 × 10^ − 3 m
=> 4.488 × 10^ − 3 m
=> 5.86 mm

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