Chapter 16, Problem 2SP

(a)

To determine

The position of first bright fringe.

Answer to Problem 2SP

The position of first bright fringe is 17.1 mm away from the center at each side.

Explanation of Solution

Given info:

Wavelength of the light is $580\text{nm}$, order of bright fringe is 1, the distance of screen from slits is $1.3\text{m}$ and spacing between adjacent slits is $0.044\text{mm}$.

Write an expression for condition of maxima.

$y=\frac{n\lambda D}{d}$

Here,

$y$ is the distance of nth bright fringe from center

$n$ is the order of the bright fringe

$D$ is the screen’s distance from slit

$\lambda$ is the wavelength

$d$ is the slit separation

Substitute $580\text{nm}$ for $\lambda$, $1$ for $n$, $1.3\text{m}$ for $D$ and $0.044\text{mm}$ for $d$ to find $y$.

$\begin{array}{c}y=\frac{\left(1\right)\left(\left(580\text{nm}\right)\left(\frac{1\text{m}}{{10}^{-9}\text{nm}}\right)\right)\left(1.3\text{m}\right)}{\left(0.044\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =17.1\times {10}^{-3}\text{m}\\ =\left(17.1\times {10}^{-3}\text{m}\right)\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =17.1\text{mm}\end{array}$

Thus, the position of first bright fringe is 17.1 mm away from the center at each side.

Conclusion:

The position of first bright fringe is 17.1 mm away from the center at each side.

(b)

To determine

The position of second bright fringe.

Answer to Problem 2SP

The position of second bright fringe is 34.3 mm away from the center at each side.

Explanation of Solution

Given info:

Wavelength of the light is $580\text{nm}$, order of bright fringe is 2, the distance of screen from slits is $1.3\text{m}$ and spacing between adjacent slits is $0.044\text{mm}$.

Write an expression for condition of maxima.

$y=\frac{n\lambda D}{d}$

Here,

$y$ is the distance of nth bright fringe from center

$n$ is the order of the bright fringe

$D$ is the screen’s distance from slit

$\lambda$ is the wavelength

$d$ is the slit separation

Substitute $580\text{nm}$ for $\lambda$, $2$ for $n$, $1.3\text{m}$ for $D$ and $0.044\text{mm}$ for $d$ to find $y$.

$\begin{array}{c}y=\frac{\left(2\right)\left(\left(580\text{nm}\right)\left(\frac{1\text{m}}{{10}^{-9}\text{nm}}\right)\right)\left(1.3\text{m}\right)}{\left(0.044\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =34.3\times {10}^{-3}\text{m}\\ =\left(34.3\times {10}^{-3}\text{m}\right)\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =34.3\text{mm}\end{array}$

Thus, the position of second bright fringe is 34.3 mm away from the center at each side.

Conclusion:

The position of second bright fringe is 34.3 mm away from the center at each side.

(c)

To determine

The position of first dark bright fringe.

Answer to Problem 2SP

The position of first dark fringe is 25.7 mm away from the center at each side.

Explanation of Solution

Given info:

Wavelength of the light is $580\text{nm}$, order of dark fringe is 1, the distance of screen from slits is $1.3\text{m}$ and spacing between adjacent slits is $0.044\text{mm}$.

Write an expression for condition of minima.

$y=\left(n+\frac{1}{2}\right)\frac{\lambda D}{d}$

Here,

$y$ is the distance of nth dark fringe from center

$n$ is the order of the bright fringe

$D$ is the screen’s distance from slit

$\lambda$ is the wavelength

$d$ is the slit separation

Substitute $580\text{nm}$ for $\lambda$, $1$ for $n$, $1.3\text{m}$ for $D$ and $0.044\text{mm}$ for $d$ to find $y$.

$\begin{array}{c}y=\left(1+\frac{1}{2}\right)\frac{\left(\left(580\text{nm}\right)\left(\frac{1\text{m}}{{10}^{-9}\text{nm}}\right)\right)\left(1.3\text{m}\right)}{\left(0.044\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =25.7\times {10}^{-3}\text{m}\\ =\left(25.7\times {10}^{-3}\text{m}\right)\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right)\\ =25.7\text{mm}\end{array}$

Thus, the position of first dark fringe is 25.7 mm away from the center at each side.

Conclusion:

The position of first dark fringe is 25.7 mm away from the center at each side.

(d)

To determine

Sketch the diffraction pattern and mark the position of the fringes.

Answer to Problem 2SP

The diffraction pattern is given in figure 1.

Explanation of Solution

Following figure gives the diffraction pattern.

Figure 1

Here, the first, second and third order bright fringes will appear at distance of 17.1 mm, 34.3 mm and 51.4 mm from center respectively at each side of the central maxima. The dark fringes of order 1, 2 and 3 will form distances 25.7 mm, 42.8 mm and 60.0 mm respectively.

Conclusion:

The diffraction pattern is given in figure 1.