A screen is placed 1.00 m behind a single slit. The central maximumin the resulting diffraction pattern on the screen is 1.60 cmwide-that is, the two first-order diffraction minima are separatedby 1.60 cm. What is the distance between the two second-orderminima?

Yasmin

Yasmin

Answered question

2021-03-22

A screen is placed 1.00 m behind a single slit. The central maximumin the resulting diffraction pattern on the screen is 1.60 cmwide-that is, the two first-order diffraction minima are separatedby 1.60 cm. What is the distance between the two second-orderminima?

Answer & Explanation

Tasneem Almond

Tasneem Almond

Skilled2021-03-24Added 91 answers

L = 1.00 m

the condition that determines the location of dark fringes in the single slit diffraction will be

sinθ;=mλ

sinθ;=mλ/WZ

θ;=sin1(mλ/W)Z

where

m=±1,±2,±3,..........

and W is the width of the slit

we know that

y=Ltanθ;

so the first order

2y1=2Ltanθ;1

0.016m=2y1

=2Ltanθ;1

=2Lθ;1
=2L(λ/W) so we get

λ/W=2y1/2L

=.............

for the second order

2y2=2Ltanθ;2
=2Ltan(sin12λ/W) solve for 2y2

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