Question

Hola mate.. ❤

the intensity at the maximum in the young's double slit experiment is Io. distance between that slits is d= 5 Labmda. Labmda is the wavelength of light used in the experiment. what will be the intensity in front of the one of the slits on the screen placed at the distance D = 10d

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Answer 1

Hey ❤Tina❤

Refer to attachment

Answer 2

Answer:

Path difference  

Δ= \frac{yd}{D}  

d=5λ

D=50λ

y= \frac{d}{2}  

= \frac{5}{2 \gamma }  

∴Δx= \frac{ \gamma }{4}  

Corresponding phase difference ϕ

ϕ=2πλΔx

=2πλ

= \frac{2π}{λ} ×λ4

= \frac{\pi}{2}  

ϕ= \frac{ϕ2}{2}  

ϕ= \frac{\pi}{4}  

í 1=I {cos}^{2}  \frac{ϕ}{2}  

= í  \frac{cos}{2}  \frac{\pi}{4}  

= í  { \frac{1}{ \sqrt{2} } }^{2}  

= \frac{i}{2}  

Hence a is the correct answer íѕ