In a double-slit experiment, what is the linear distance on the screen between adjacent maxima if the wavelength is \(546 \mathrm{nm}\), the slit separation is \(0.100 \mathrm{mm},\) and the slit-screen separation is \(20.0 \mathrm{cm} ?\)

Short Answer

Expert verified
Question: Determine the linear distance between adjacent maxima on the screen in a double-slit experiment, given a wavelength of 546 nm, a slit separation of 0.100 mm, and a slit-screen separation of 20.0 cm. Answer: The linear distance between adjacent maxima in this double-slit experiment is 2.184 mm.

Step by step solution

01

First, we need to convert the given values into meters for a consistent unit. The wavelength is given in \(nm\), slit separation in \(mm\), and slit-screen separation is given in \(cm\), so let's convert them: Wavelength (\(\lambda\)) = \(546 \times 10^{-9}\) meters Slit separation (\(d\)) = \(0.100 \times 10^{-3}\) meters Slit-screen separation (\(L\)) = \(20.0 \times 10^{-2}\) meters #Step 2: Apply the formula for maxima positions#

Next, we apply the formula \(y = m\frac{L\lambda}{d}\) to find the positions of two consecutive maxima, for example \(m = 1\) and \(m = 2\). For \(m = 1\): \(y_1 = 1 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) For \(m = 2\): \(y_2 = 2 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) #Step 3: Calculate the linear distance between adjacent maxima#
02

To find the linear distance between adjacent maxima, we need to find the difference between \(y_1\) and \(y_2\). Linear Distance (\(\Delta y\)) = \(y_2 - y_1\) Substitute the expressions for \(y_1\) and \(y_2\): \(\Delta y = 2 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})} - 1 \times \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) Simplify the expression: \(\Delta y = \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})}\) #Step 4: Find the numerical value of the linear distance#

Now, we can calculate the linear distance by substituting the values and evaluating the expression: \(\Delta y = \frac{(20.0 \times 10^{-2}) (546 \times 10^{-9})}{(0.100 \times 10^{-3})} = 0.002184\) meters Finally, convert the result to millimeters for better readability: Linear Distance (\(\Delta y\)) = \(2.184\) mm

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two slits separated by \(20.0 \mu \mathrm{m}\) are illuminated by light of wavelength \(0.50 \mu \mathrm{m} .\) If the screen is \(8.0 \mathrm{m}\) from the slits, what is the distance between the \(m=0\) and \(m=1\) bright fringes?
In bright light, the pupils of the eyes of a cat narrow to a vertical slit \(0.30 \mathrm{mm}\) across. Suppose that a cat is looking at two mice $18 \mathrm{m}$ away. What is the smallest distance between the mice for which the cat can tell that there are two mice rather than one using light of $560 \mathrm{nm} ?$ Assume the resolution is limited by diffraction only.
Diffraction and the Resolution of Optical Instruments The Hubble Space Telescope (HST) has excellent resolving power because there is no atmospheric distortion of the light. Its 2.4 -m-diameter primary mirror can collect light from distant galaxies that formed early in the history of the universe. How far apart can two galaxies be from each other if they are 10 billion light-years away from Earth and are barely resolved by the HST using visible light with a wavelength of \(400 \mathrm{nm} ?\)
A double slit is illuminated with monochromatic light of wavelength $600.0 \mathrm{nm} .\( The \)m=0\( and \)m=1\( bright fringes are separated by \)3.0 \mathrm{mm}\( on a screen \)40.0 \mathrm{cm}$ away from the slits. What is the separation between the slits? [Hint: Is the small angle approximation justified?]
Sonya is designing a diffraction experiment for her students. She has a laser that emits light of wavelength \(627 \mathrm{nm}\) and a grating with a distance of \(2.40 \times 10^{-3} \mathrm{mm}\) between slits. She hopes to shine the light through the grating and display a total of nine interference maxima on a screen. She finds that no matter how she arranges her setup, she can see only seven maxima. Assuming that the intensity of the light is not the problem, why can't Sonya display the \(m=4\) interference maxima on either side?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free