Chapter 24, Problem 14P

Monochromatic light of wavelength λ is incident on a pair of slits separated by 2.40 × 10⁻⁴m. and forms an interference pattern on a screen placed 1.80 m away from the slits. The first-order bright fringe is 4.52 mm from the center of the central maximum. (a) Draw a picture, labeling the angle θ and the legs of the right triangle associated with the first-order bright fringe. (b) Compute the tangent of the angle θ associated with the first-order bright fringe. (c) Find the angle corresponding to the first-order bright fringe and compute the sine of that angle. Are the sine and tangent of the angle comparable in value? Does your answer always hold true? (d) Calculate the wavelength of the light. (e) Compute the angle of the fifth-order bright fringe. (f) Find its position on the screen.

To determine

Draw the picture of the bright fringe formation.

Answer to Problem 14P

Following figure shows the first bright fringe formation.

Explanation of Solution

Following figure shows the first bright fringe formation.

Here,

$d$ is the slit separation

$L$ is the screen distance

${\theta }_1$ is the angle associated with first bright fringe

$y_1$ is the distance between the central maxima and the first bright fringe

Conclusion:

Following figure shows the first bright fringe formation.

To determine

The tangent of the angle associated with the first order bright fringe.

Answer to Problem 14P

The tangent of the angle associated with the first order bright fringe is $2.51\times {10}^{-3}$.

Explanation of Solution

Given info:

The distance between the central maxima and the first bright fringe is $4.52\text{mm}$.

The screen separation is $1.80\text{m}$.

Explanation:

Formula to calculate the tangent of the angle associated with the first order bright fringe is,

$\tan {\theta }_1=\frac{y_1}{L}$

Substitute $4.52\text{mm}$ for $y_1$ and $1.80\text{m}$ for $L$ to find $\tan {\theta }_1$.

$\begin{array}{c}\tan {\theta }_1=\frac{\left(4.52\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{1.80\text{m}}\\ =2.51\times {10}^{-3}\end{array}$

Conclusion:

The tangent of the angle associated with the first order bright fringe is $2.51\times {10}^{-3}$.

To determine

The angle associated with the first fringe and the sine of the angle. The comparability of sine and tangent of the angle.

Answer to Problem 14P

The angle associated with the first fringe is $0.144°$ and the sine of the angle is $2.51\times {10}^{-3}$. The sine and the tangent of the angle are nearly same since the angle is very small. For larger angle the sine and tangent will not be comparable.

Explanation of Solution

Formula to calculate the angle associated with the first order bright fringe is,

${\theta }_1={\tan }^{-1}\left(\tan {\theta }_1\right)$

Substitute $2.51\times {10}^{-3}$ for $\tan {\theta }_1$ to find ${\theta }_1$.

$\begin{array}{c}{\theta }_1={\tan }^{-1}\left(2.51\times {10}^{-3}\right)\\ =0.144°\end{array}$

Formula to calculate the sine of the angle associated with the first order bright fringe is,

$\sin {\theta }_1=\sin \left({\tan }^{-1}\left(\tan {\theta }_1\right)\right)$

Substitute $0.144°$ for ${\tan }^{-1}\left(\tan {\theta }_1\right)$ to find $\sin {\theta }_1$.

$\begin{array}{c}{\theta }_1=\sin \left(0.144°\right)\\ =2.51\times {10}^{-3}\end{array}$

The sine and the tangent of the angle are nearly same since the angle is very small. For larger angle the sine and tangent will not be comparable.

Conclusion:

The angle associated with the first fringe is $0.144°$ and the sine of the angle is $2.51\times {10}^{-3}$. The sine and the tangent of the angle are nearly same since the angle is very small. For larger angle the sine and tangent will not be comparable.

To determine

The wavelength of the light.

Answer to Problem 14P

The wavelength of the light is $603\text{nm}$.

Explanation of Solution

Given info:

The slit separation is $2.40\times {10}^{-4}\text{m}$.

The order of first bright fringe is $1$.

Explanation:

Formula to calculate the wavelength is,

$\lambda =\frac{d\sin {\theta }_1}{m}$

Here,

$\lambda$ is the wavelength

$m$ is the order of first bright fringe

Substitute $2.40\times {10}^{-4}\text{m}$ for $d$, $1$ for $m$ and $0.144°$ for ${\theta }_1$ to find $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(2.40\times {10}^{-4}\text{m}\right)\sin \left(0.144°\right)}{1}\\ =603\times {10}^{-9}\text{m}\\ \text{=}\left(603\times {10}^{-9}\text{m}\right)\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =603\text{nm}\end{array}$

Conclusion:

The wavelength of the light is $603\text{nm}$.

To determine

The angle of the fifth order bright fringe.

Answer to Problem 14P

The angle of the fifth order bright fringe is $0.720°$.

Explanation of Solution

Given info:

The order of fifth bright fringe is $5$.

Explanation:

Formula to calculate the angle of fifth order bright fringe is,

${\theta }_5={\sin }^{-1}\left(\frac{m_5\lambda }{d}\right)$

${\theta }_5$ is the angle of the fifth order bright fringe

$m_5$ is the order of fifth bright fringe

Substitute $2.40\times {10}^{-4}\text{m}$ for $d$, $5$ for $m$ and $603\text{nm}$ for $\lambda$ to find ${\theta }_5$.

$\begin{array}{c}{\theta }_5={\sin }^{-1}\left(\frac{5\left(603\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(2.40\times {10}^{-4}\text{m}\right)}\right)\\ =0.720°\end{array}$

Conclusion:

The angle of the fifth order bright fringe is $0.720°$.

To determine

The position of the fifth bright fringe on the screen.

Answer to Problem 14P

The position of the fifth bright fringe on the screen is $2.26\text{cm}$.

Explanation of Solution

Formula to calculate the position of the fifth bright fringe is,

$y_5=L\tan {\theta }_5$

Here,

$y_5$ is the position of the fifth bright fringe on the screen

Substitute $1.80\text{m}$ for $L$ and $0.720°$ for ${\theta }_5$ to find $y_5$.

$\begin{array}{c}{y}_5=\left(1.80\text{m}\right)\tan \left(0.720°\right)\\ =2.26\times {10}^{-2}\text{m}\\ =\left(2.26\times {10}^{-2}\text{m}\right)\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)\\ =2.26\text{cm}\end{array}$

Conclusion:

The position of the fifth bright fringe on the screen is $2.26\text{cm}$.