91Ó°ÊÓ

Light of wavelength \(410 \mathrm{nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(1.2 \times 10^{-5} \mathrm{m} .\) The distance between the grating and the viewing screen is \(0.15 \mathrm{m}\). A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (Hint: The diffraction angles are small enough that the approximation \(\tan \theta=\sin \theta\) can be used. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right),\) what is the distance \(y ?\)

Step by step solution

01

Calculate the angle for the second-order bright fringe in vacuum

The formula for the diffraction condition is given by \( d \sin \theta = m \lambda \), where \( d \) is the slit separation, \( m \) is the order number, and \( \lambda \) is the wavelength. For the second-order bright fringe \( (m = 2) \) and \( \lambda = 410 \, \mathrm{nm} = 410 \times 10^{-9} \, \mathrm{m} \).Thus, \( \sin \theta = \frac{2 \times 410 \times 10^{-9}}{1.2 \times 10^{-5}} \). Calculate this to find \( \sin \theta \).

02

Use the small angle approximation to find the second-order distance in vacuum

Given \( \tan \theta \approx \sin \theta \) for small angles, and the viewing distance \( L = 0.15 \, \mathrm{m} \), we have \( y = L \tan \theta \).Use the \( \tan \theta = \sin \theta \) approximation to find the distance \( y = L \sin \theta \). Substitute the result from Step 1 to calculate \( y \).

03

Adjust wavelength for water medium

When the setup is submerged in water, the effective wavelength \( \lambda_{\text{water}} = \frac{\lambda}{n_{\text{water}}} = \frac{410 \times 10^{-9}}{1.33} \). Calculate \( \lambda_{\text{water}} \).

04

Calculate the second-order distance in water

Repeat Steps 1 and 2 using the adjusted \( \lambda_{\text{water}} \).- Use the diffraction formula \( d \sin \theta_{\text{water}} = m \lambda_{\text{water}} \) to find \( \sin \theta_{\text{water}} \).- Calculate \( y_{\text{water}} = L \sin \theta_{\text{water}} \) using the new angle, and finally compute \( y_{\text{water}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength in Vacuum

When we talk about the wavelength of light, we're referring to the distance between successive peaks of a light wave. The wavelength is crucial when dealing with diffraction gratings. In a vacuum, the speed of light is at its maximum, allowing wavelengths to be measured without interference from other substances. Light in a vacuum has a refractive index of 1, meaning no bending occurs. For example, in our exercise, the given wavelength of light is 410 nm. This is a common measurement used in optics. This precise measurement is needed to calculate angles and positions of bright fringes in diffraction patterns.

Small Angle Approximation

The small angle approximation is a handy mathematical tool. It is particularly useful in diffraction problems where angles are tiny. For small angles, the values of sine and tangent are almost the same. This simplifies calculations. The approximation is expressed as: \[\tan \theta \approx \sin \theta \approx \theta \]This means you can replace \(\tan \theta \) with \(\sin \theta \), which simplifies many steps, like finding distances on a screen. For example, when calculating the angle for a second-order bright fringe, using \( \tan \theta = \sin \theta \) allows you to find distances by simple multiplication.

Refractive Index

The refractive index tells us how much light bends or slows down when entering a medium. It's a ratio comparing the speed of light in a vacuum to its speed in another material. For instance, water has a refractive index of 1.33, meaning light travels slower in water than in a vacuum. In our exercise, when the apparatus is submerged in water, the wavelength in water becomes: \[\lambda_{\text{water}} = \frac{\lambda}{n_{\text{water}}}\]This change affects how light spreads out, altering the angles of diffraction and thus the position of bright fringes.

Bright Fringe Order

Bright fringes appear on a screen when light waves interfere constructively. In a diffraction grating, these fringes are identified by an order number \(m\). The higher the order, the further from the center the fringe appears.The formula for finding the position of a bright fringe involves the order number:\[d \sin \theta = m \lambda\]Here, \(d\) is the slit separation, \(\theta\) the angle, and \(\lambda\) the wavelength. For second-order fringes, \(m = 2\). This calculation helps determine the distance of these bright spots from the central fringe. By understanding this, you can predict where light will form patterns on a screen.