Question

A 0.50-mm-diameter hole is illuminated by light of wavelength 510 nm. What is the width of the central maximum on a screen 2.2 m behind the slit (in mm

Answer 1

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  × $10 ^{− 9 }m$

The angle subtended by the maxima is given by  

sin θ  =  m λ/ d  

Where m = 1,2,3,4..... for the positions of different maxima

The angle subtended by the first central maxima is given by  

sin θ  =  1 × λ /d  

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

=>   1.020 × 10 − 3  

Now from the diagram, we can see  

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 total width of the central maximum in the diffraction pattern on the screen  

= > 2 Y  =  2  ×  2.93 × 10^ − 3 m

=> 4.488  ×  10^ − 3 m

=>  5.86  mm