**Answer :**

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

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 hole is illuminated by a light of wavelength **

**510 n m = 510 × **

**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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