## 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:

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?

### Answer:

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

## Answer ( 1 )

Answer Above.