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You illuminate a slit with a width of 70.3 μm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 2.11 m from the slit. What is the width, in centimeters, of the pattern?

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Answer:

4.3 cm

Step-by-step explanation:

We are given that

Width,d=70.3
\mu m=70.3*10^(-6) m


1\mu m=10^(-6) m

Wavelength,
\lambda=719 nm=719* 10^(-9) m


1nm=10^(-9) m


r=2.11 m

We have to find the width in cm of the pattern.

The angle for the first minimum m=1


sin\theta_(min)\approx \theta=(\lambda)/(d)=(719* 10^(-9))/(70.3* 10^(-6))=0.0102 rad


y=r\theta=2.11* 0.0102=0.0215 m

The width of the pattern=
2y=2* 0.0215=0.043 m=0.043* 100=4.3 cm

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