Wave Optics: Doppler Effect

The fractional change in frequency is given by Δν/ν = -v_radial/c, where v_radial is the component of the source's velocity along the line joining the observer to the source.

The Doppler effect allows astronomers to measure the radial velocities of distant galaxies based on the shift in light frequency.

Using the relation Δλ/λ = Δν/ν, we can determine v_radial = Δλ/λ * c to find the speed of the galaxy.

<p>The frequency of both reflected and refracted light remains the same as that of the incident light.</p> Signup and view all the answers

<p>In the wave picture, the intensity of light is proportional to the square of the amplitude of the wave.</p> Signup and view all the answers

<p>The speed of light decreases when it travels from a rarer medium to a denser medium.</p> Signup and view all the answers

<p>In the photon picture, the intensity of light is determined by the number of photons crossing a unit area per unit time.</p> Signup and view all the answers

<p>The Doppler effect formula is modified for lightwaves at high speeds to account for the effects of Einstein's special theory of relativity.</p> Signup and view all the answers

<p>Replacing S₂P + S₁P with 2D simplifies calculations and introduces negligible error in the approximation when $d$ and $x$ are much smaller than $D$.</p> Signup and view all the answers

<p>Constructive interference occurs when $x = x_n = ηλD/d$, where $n$ is an integer (0, ±1, ±2, ...).</p> Signup and view all the answers

<p>The fringe width $β$ indicates the distance between consecutive bright and dark fringes, providing insight into the characteristics of the interference pattern.</p> Signup and view all the answers

<p>The fringe pattern produced is ideally a hyperbola, but appears as nearly straight lines when the distance $D$ is large compared to the fringe width.</p> Signup and view all the answers

<p>Whenever $S₂P - S₁P$ is an integral multiple of $ ext{λ}$, the fringe is bright; if it is an odd integral multiple of $ ext{λ/2}$, the fringe is dark.</p> Signup and view all the answers

<p>Diffraction causes alternate dark and bright regions around the geometrical shadow of an opaque object, similar to interference patterns.</p> Signup and view all the answers

<p>Point O is bright because it is equidistant from S₁ and S₂, resulting in equal path lengths and thus constructive interference.</p> Signup and view all the answers

<p>Diffraction is a general characteristic that occurs in all types of waves, including sound, light, water, and matter waves.</p> Signup and view all the answers

<p>Diffraction is the spreading out of light waves as they pass through narrow slits, allowing light to turn corners and enter shadowed areas.</p> Signup and view all the answers

<p>The intensity pattern shows a broad central bright region with alternating dark and bright regions on either side, diminishing in intensity as they move away from the center.</p> Signup and view all the answers

<p>A monochromatic light source ensures that all light waves are of the same wavelength, leading to clear and distinct interference patterns.</p> Signup and view all the answers

<p>The angle θ represents the angle between the normal to the slit plane and the line connecting a point on the screen to a point in the slit.</p> Signup and view all the answers

<p>The path difference NP - LP is equal to $a sin(θ)$, which can be approximated as $aθ$ for small angles.</p> Signup and view all the answers

<p>Fresnel used integral calculus to calculate the contributions of multiple secondary sources within the slit to the resultant intensity at a point on the screen.</p> Signup and view all the answers

<p>An increased number of coherent sources results in a clearer and more defined diffraction pattern due to the cumulative effects of interference.</p> Signup and view all the answers

<p>Phase differences cause some waves to interfere constructively while others interfere destructively, shaping the overall intensity distribution of the diffraction pattern.</p> Signup and view all the answers

<p>Feynman suggests that the distinction lies in the number of sources involved; interference typically involves a few sources, while diffraction often involves many sources.</p> Signup and view all the answers

<p>The observed pattern on the screen is a combination of single-slit diffraction from each slit and the double-slit interference pattern.</p> Signup and view all the answers

<p>The number of interference fringes is influenced by the ratio $d/a$, where $d$ is the distance between the slits and $a$ is the width of a slit.</p> Signup and view all the answers

<p>Diffraction causes the beam of light, rather than focusing to a point, to be focused to a spot of finite area.</p> Signup and view all the answers

<p>The primary purpose of the eyepiece is to provide magnification of the image produced by the telescope's objective.</p> Signup and view all the answers

<p>Stars that are not resolved cannot be distinguished further because mechanical magnification does not improve the resolution of the image.</p> Signup and view all the answers

<p>The lens must be well corrected for aberrations to focus a parallel beam of light accurately.</p> Signup and view all the answers

<p>The analysis of the diffraction pattern can be approached by considering a plane wave incident on a circular aperture followed by a convex lens.</p> Signup and view all the answers

<p>The radius is approximately given by $r_0 ≈ 0.61 \frac{\lambda f}{a}$.</p> Signup and view all the answers

<p>A larger diameter objective lens decreases $\Delta \theta$, allowing better resolution of two stars.</p> Signup and view all the answers

<p>The limit of resolution is approximately $2.9 \times 10^{-7}$ radians.</p> Signup and view all the answers

<p>The radius of the central bright spot depends on the focal length $f$ and the diameter $2a$ of the aperture.</p> Signup and view all the answers

<p>By observing black and white stripes with increasing widths, you can gauge resolving power.</p> Signup and view all the answers

<p>It implies that a larger aperture size $a$ leads to a smaller $\Delta \theta$, enhancing resolution.</p> Signup and view all the answers

<p>It determines the limit of resolution, affecting the clarity of images produced by telescopes or microscopes.</p> Signup and view all the answers

<p>The approximate radius $r_0$ is around $1.2 \mu m$.</p> Signup and view all the answers

<p>A larger diameter objective increases the telescope's resolving power, allowing for better resolution of stars.</p> Signup and view all the answers

<p>The limit of resolution is approximately $2.9 imes 10^{-7}$ radians.</p> Signup and view all the answers

<p>A shorter wavelength leads to a smaller $Δθ$, resulting in better resolving power and the ability to distinguish finer details.</p> Signup and view all the answers

<p>Similar to telescopes, a larger diameter objective lens in a microscope enhances its resolving power, allowing better differentiation between small details.</p> Signup and view all the answers

<p>You can use black and white stripes of equal width and measure the point at which the black stripes merge into one another.</p> Signup and view all the answers

<p>Record the width of the separating white stripe and the distance from the wall to your eye, then compute $d/D$ for resolution.</p> Signup and view all the answers

<p>The angle helps relate the size of the objective lens to the distance required for achieving a particular resolving power.</p> Signup and view all the answers

<p>By knowing the resolution of your eye and the distance to the speck, you can use this information to estimate its size.</p> Signup and view all the answers