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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 661: Lesson Check

Exercise 37

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In order for diffraction to happen a wave must “bump into” an obstacle or a slit whose dimensions are comparable to its’ wavelength. Since the wavelength of light is of the order of $10^{-7}text{ m}$ and the wavelength of sound is of the order of a few dozens of centimeter it is much easier to achieve more noticeable diffraction of sound than of light.

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Exercise 38

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The light that is coming to us from them diffracts on our pupils and they become blurred together.

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Exercise 39

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We call this simply resolution.

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Resolution.

Exercise 40

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The smaller the angle of diffraction, the less blurring and so the greater the resolution. Since the angle of diffraction is obtained from

$$
sintheta=1.22frac{lambda}{D}
$$
we see that by increasing $D$ we decrease the sine and so the angle of diffraction (since the sine is an increasing function in the range of $0-90^circ$) which enhances camera’s resolution.

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Exercise 41

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$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
lambda &= 676 text{ nm} \
W &= 7.64 ~ mu text{m}
end{align*}
$$

In order to calculate the angle to the first dark fringe above the central fringe, we need to apply the expression for dark fringes in single-slit interference. The value of the integer, $m$ , is equal to 1 since it is the first dark fringe.

$$
begin{align*}
W sin theta &= m lambda \
sin theta &= dfrac{m lambda}{W} \
theta &= {sin}^{-1} left( dfrac{m lambda}{W} right) \
&= {sin}^{-1} left[ dfrac{1 (676 cdot 10^{-9} text{ m})}{7.64 cdot 10^{-6} text{ m}} right]
end{align*}
$$

$$
{boxed{theta = 5.078 text{textdegree}}}
$$

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$$
{theta = 5.078 text{textdegree}}
$$

Exercise 42

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$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
W &= 2.7 ~ mu text{m} \
theta &= 12 text{textdegree}
end{align*}
$$

The wavelength of the light that passes through a given slit width to the first dark fringe with integer equal to 1 can be obtain by applying the expression for dark fringes in single-slit interference :

$$
begin{align*}
W sin theta &= m lambda \
lambda &= dfrac{W sin theta}{m} \
&= dfrac{2.7 cdot 10^{-6} text{ m} (sin 12 text{textdegree})}{1}
end{align*}
$$

$$
{boxed{lambda = 561.4 text{ nm}}}
$$

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$$
{lambda = 561.4 text{ nm}}
$$

Exercise 43

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$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
lambda &= 592 text{ nm} \
theta &= 21 text{textdegree}
end{align*}
$$

The slit width, $W$ , can be obtain by applying the relation for dark fringes in single-slit interference :

$$
begin{align*}
W sin theta &= m lambda \
W &= dfrac{m lambda}{sin theta} \
&= dfrac{1 (592 text{ nm})}{sin 21 text{textdegree}}
end{align*}
$$

$$
{boxed{W = 1651.93 text{ nm}}}
$$

Result

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$$
{W = 1651.93 text{ nm}}
$$

Exercise 44

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$textbf{Given values:}$

$$
begin{align*}
m &= 1 \
lambda_{He} &= 632.8 text{ nm} \
theta &= 17 text{textdegree} \
lambda_Y &= 591 text{ nm}
end{align*}
$$

$textbf{(a)}$ The width of the slit, $W$ , produced by the passing light from a helium-neon laser can be obtain by applying the relation for dark fringes in single-slit interference :

$$
begin{align*}
W sin theta &= m lambda_{He} \
W &= dfrac{m lambda_{He}}{sin theta} \
&= dfrac{1 (632.8 text{ nm})}{sin 17 text{textdegree}}
end{align*}
$$

$$
{boxed{W = 2164.37 text{ nm}}}
$$

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$textbf{(b)}$ If a monochromatic yellow light of wavelength $591 text{ nm}$ is used instead of the helium-neon laser, the angle to the first dark fringe will be $textbf{less than}$ to $17 text{textdegree}$. From the expression for dark fringes in single-slit interference, the angle is directly proportional with the wavelength therefore, the smaller the wavelength is the lesser the angle will be.

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$textbf{(a)}$ ${W = 2164.37 text{ nm}}$

$textbf{(b)}$ $textbf{less than}$ to $17 text{textdegree}$

Exercise 45

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$textbf{Given values:}$

$$
begin{align*}
lambda &= 690 text{ nm} \
D &= 5.1 text{ mm}
end{align*}
$$

The minimum angle of separation between the two point sources of light can be calculated using the expression for the diffraction pattern of a circular opening :

$$
begin{align*}
sin theta &= 1.22 ~ dfrac{lambda}{D} \
theta &= {sin}^{-1} left( 1.22 ~ dfrac{lambda}{D} right) \
&= {sin}^{-1} left( 1.22 times dfrac{690 cdot 10^{-9} text{ m}}{0.0051 text{ m}}right)
end{align*}
$$

$$
{boxed{theta = 0.00946 text{textdegree}}}
$$

Result

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$$
{theta = 0.00946 text{textdegree}}
$$

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