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Page 646: Practice Problems

Exercise 6

Step 1

1 of 2

a) if the the wavelength increases so does the spacing between the fringes and thus the linear distance to the tenth fringe also increases.

b) From the diffraction condition for tenth bright fringe ($m=10$) we get

$$
10lambda = dsintheta_{10}Rightarrow sintheta_{10}=frac{10lambda}{d}Rightarrowtheta_{10}=arcsinleft(frac{10lambda}{d}right) = 3.71^circ.
$$
The geometry here remains the same so if $y$ is the required distance we get

$$
y=Ltantheta_{10} = 2.3text{ m}times tan3.71^circ =15text{ cm}.
$$
Which, by being greater than $12text{ cm}$ confirms our prediction.

Result

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Exercise 7

Step 1

1 of 2

Let us first write down the diffraction condition for the 3rd bright fringe ($m=3$)

$$
3lambda = dsinthetaRightarrow 3lambda = 0.0334times10^{-3}text{ m }timessin3.21^circ
$$
and this yields

$$
lambda = frac{0.0334times10^{-3}text{ m }timessin3.21^circ}{3} = 623text{ nm}.
$$

Result

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Exercise 8

Step 1

1 of 2

Let us first write down the diffraction condition for the 1st bright fringe ($m=1$)

$$
lambda=dsinthetaRightarrow d=frac{lambda}{sintheta}=frac{520times10^{-9}text{ m}}{sin0.21^circ} = 2.5text{ $mu$m}.
$$

Result

2 of 2

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Page 646: Practice Problems

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