Chapter 36, Problem 14P

Monochromatic light of wavelength λ is incident on a pair of slits separated by 2.40 × 10⁻⁴ m and forms an interference pattern on a screen placed 1.80 m from the slits. The first-order bright fringe is at a position y_bright = 4.52 mm measured from the center of the central maximum. From this information, we wish to predict where the fringe for n = 50 would be located. (a) Assuming the fringes are laid out linearly along the screen, find the position of the n = 50 fringe by multiplying the position of the n = 1 fringe by 50.0. (b) Find the tangent of the angle the first-order bright fringe makes with respect to the line extending from the point midway between the slits to the center of the central maximum. (c) Using the result of part (b) and Equation 36.2, calculate the wavelength of the light. (d) Compute the angle for the 50th-order bright fringe from Equation 36.2. (e) Find the position of the 50th-order bright fringe on the screen from Equation 36.5. (f) Comment on the agreement between the answers to parts (a) and (e).

To determine

The position of the ${50}^{\text{th}}$ fringe.

Answer to Problem 14P

The position of the ${50}^{\text{th}}$ fringe is $22.6\text{cm}$.

Explanation of Solution

Given info: The separation of the slit is $2.40\times {10}^{-4}\text{m}$, the distance between the screen and the slit is $1.80\text{m}$ and the position of the first-order bright fringe is $y_{\text{bright}}=4.52\text{mm}$.

The formula to calculate the position of the ${50}^{\text{th}}$ fringe is,

$y=50{\left(y_{\text{bright}}\right)}_{m=1}$

Here,

$y_{\text{bright}}$ is the position of the first-order bright fringe.

Substitute $4.52\text{mm}$ for $y_{\text{bright}}$ in above equation to find the value of $y$.

$\begin{array}{c}y=50\left(4.52\text{mm}\right)\\ =22.6\text{cm}\end{array}$

Conclusion:

Therefore, the position of the ${50}^{\text{th}}$ fringe is $22.6\text{cm}$.

To determine

The tangent of the angle of the first-order bright fringe with respect to the point midway between the slits to the center of the central maximum.

Answer to Problem 14P

The tangent of the angle of the first-order bright fringe with respect to the point midway between the slits to the center of the central maximum is $2.51\times {10}^{-3}$.

Explanation of Solution

Given info: The separation of the slit is $2.40\times {10}^{-4}\text{m}$, the distance between the screen and the slit is $1.80\text{m}$ and the position of the first-order bright fringe is $y_{\text{bright}}=4.52\text{mm}$.

The formula to calculate the tangent of the angle is,

$\tan {\theta }_1=\frac{{\left(y_{\text{bright}}\right)}_{m=1}}{L}$

Here,

$L$ is the distance between the screen and the slit.

Substitute $4.52\text{mm}$ for ${\left(y_{\text{bright}}\right)}_{m=1}$ and $1.80\text{m}$ for $L$ in the above equation.

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

Conclusion:

Therefore, the tangent of the angle of the first-order bright fringe with respect to the point midway between the slits to the center of the central maximum is $2.51\times {10}^{-3}$.

To determine

The wavelength of the light.

Answer to Problem 14P

The wavelength of the light is $6.03\times {10}^{-7}\text{m}$.

Explanation of Solution

Given info: The separation of the slit is $2.40\times {10}^{-4}\text{m}$, the distance between the screen and the slit is $1.80\text{m}$ and the position of the first-order bright fringe is $y_{\text{bright}}=4.52\text{mm}$.

The formula to calculate the tangent of the angle is,

$\tan {\theta }_1=\frac{{\left(y_{\text{bright}}\right)}_{m=1}}{L}$

Substitute $4.52\text{mm}$ for ${\left(y_{\text{bright}}\right)}_{m=1}$ and $1.80\text{m}$ for $L$ in the above equation.

$\begin{array}{c}\tan {\theta }_1=\frac{4.52\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{m}}}{1.80\text{m}}\\ \tan {\theta }_1=2.51\times {10}^{-3}\\ {\theta }_1=0.144°\end{array}$]

The formula to calculate the wavelength is,

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

Here,

$m$ is the number of the maxima.

$\lambda$ is the wavelength of the light.

Substitute $1$ for $m$, $2.40\times {10}^{-4}\text{m}$ for $d$ and $0.144°$ for ${\theta }_1$ in above equation.

$\begin{array}{c}\lambda =\left(2.40\times {10}^{-4}\text{m}\right)\sin \left(0.144°\right)\\ =6.03\times {10}^{-7}\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the light is $6.03\times {10}^{-7}\text{m}$.

To determine

The angle of the ${50}^{\text{th}}$ order-bright fringe.

Answer to Problem 14P

The angle of the ${50}^{\text{th}}$ order-bright fringe is $7.21°$.

Explanation of Solution

Given info: The separation of the slit is $2.40\times {10}^{-4}\text{m}$, the distance between the screen and the slit is $1.80\text{m}$ and the position of the first-order bright fringe is $y_{\text{bright}}=4.52\text{mm}$.

The formula to calculate the angle of the ${50}^{\text{th}}$ order-bright fringe is,

${\theta }_{50}={\sin }^{-1}\left(50\sin {\theta }_1\right)$

Substitute $0.144°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\theta }_{50}={\sin }^{-1}\left(50\sin \left(0.144°\right)\right)\\ =7.21°\end{array}$

Conclusion:

Therefore, the angle of the ${50}^{\text{th}}$ order-bright fringe is $7.21°$.

To determine

The position of the ${50}^{\text{th}}$ order-bright fringe.

Answer to Problem 14P

The position of the ${50}^{\text{th}}$ order-bright fringe is $2.28\text{cm}$.

Explanation of Solution

Given info: The separation of the slit is $2.40\times {10}^{-4}\text{m}$, the distance between the screen and the slit is $1.80\text{m}$ and the position of the first-order bright fringe is $y_{\text{bright}}=4.52\text{mm}$.

The formula to calculate the position of the ${50}^{\text{th}}$ order-bright fringe is,

$y_{50}=L\tan {\theta }_{50}$

Substitute $1.80\text{m}$ for $L$ and $7.21°$ for ${\theta }_{50}$ in above equation to find the value of $y_{50}$.

$\begin{array}{c}{y}_{50}=\left(1.80\text{m}\right)\tan \left(7.21°\right)\\ =2.28\times {10}^{-2}\text{m}\times \frac{{10}^2\text{cm}}{1\text{m}}\\ =2.28\text{cm}\end{array}$

Conclusion:

Therefore, the position of the ${50}^{\text{th}}$ order-bright fringe is $2.28\text{cm}$.

To determine

The comment on the agreement between the answers to parts $\left(a\right)$ and part $\left(e\right)$.

Answer to Problem 14P

The answer to part $\left(a\right)$ and part $\left(e\right)$ are very close bet not equal as the fringes are not laid linear.

Explanation of Solution

Given info: The separation of the slit is $2.40\times {10}^{-4}\text{m}$, the distance between the screen and the slit is $1.80\text{m}$ and the position of the first-order bright fringe is $y_{\text{bright}}=4.52\text{mm}$.

The difference in the position of the ${50}^{\text{th}}$ order-bright fringe is different as calculated in part $\left(a\right)$ and part $\left(e\right)$ so it can be deduced from the results that the fringes are not laid out linearly on the screen as assumed in part $\left(a\right)$ and the nonlinearity is evident for very large angles.

Conclusion:

Therefore, the answer to part $\left(a\right)$ and part $\left(e\right)$ are very close bet not equal as the fringes are not laid linear.