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Optics and Wave Phenomena

Problem 1

Unpolarized light is incident on a pair of ideal linear polarizers whose transmission axes make an angle of 45 degrees with each other. What percentage of the incident intensity is the transmitted light intensity through both polarizers?

Solution 1

Answer

Problem 2

During a hurricane, a 1200Hz warning siren on the town hall sounds. The wind is blowing at 55m/s in a direction from the siren toward a person 1 km away. With what frequency does the sound wave reach the person? (The speed of sound in air is 330m/s)

Solution 2

Answer

Problem 3

Sound waves moving at 350 m/s diffract out of a speaker enclosure with an opening tha tis a long rectangular slid 0.14 m across. At about what frequency will the sound first disappear at an angle of 45 ° from the normal to the speaker face?

Solution 3

Answer

Problem 4

An organ pipe closed at one end and open at the other, is designed to have a fundamental frequency of C (131Hz). What is the frequency of the next higher harmonic for this pipe?

Solution 4

Answer

Problem 5

X-rays of wavelength λ = 0.250nm are incident on the face of a crystal at angle θ, measured from the crystal surface. The smallest angle that yeilds an intense reflected beam is θ = 14.5° . How many nanometers is the value of the interplanar spacing d? (sin 14.5° = 1/4)

Solution 5

Answer

Problem 6

574

The figure above shows an object O placed at a distance R to the left of a convex spherical mirror that has a radius of curvature R. Point C is the center of curvature of the mirror. The image formed by the mirror is at

Solution 6

Answer

Problem 7

A uniform thin film of soapy water with index of refraction n = 1.33 is viewed in air via reflected light. The film appears dark for long wavelengths and first appears bright for λ = 540 nm. What is the next shorter wavelength (in nm) at which the film will appear bright on reflection?

Solution 7

Answer

Problem 8

576

A model of an optical fiber is shown in the figure above. The optical fiber has an index of refraction n and is surrounded by free space. What angles of incidence θ will result in the light staying in the optical fiber?

  1. θ > sin-1 √(n2 - 1)
  2. θ > sin-1 √(n2 + 1)
  3. θ < sin-1 √(n2 - 1)
  4. sin-1 √(n2 - 1) < θ < sin-1 √(n2 + 1)
  5. θ < sin-1 √(n2 + 1)

Solution 8

Answer

Problem 9

697

A beam of light has a small wavelength spread δλ about a central wavelength λ. The beam travels in vacuum until it enters a glass plate at an angle θ relative to the normal to the plate, as shown in the figure above. The index of refraction of the glass is given by n(λ). What is the angular spread δθ' of the refracted beam?

Solution 9

Answer

Problem 10

y = A \mathrm{sin} 2 \pi (\frac{1}{T} - \frac{x}{\lambda})

The equation above, where A, T, and λ are positive constants, represents a wave whose

  1. velocity is in the negative x-direction
  2. period is T / λ
  3. speed is x / t
  4. amplitude is 2A
  5. speed is λ / T

Solution 10

The original equation was
y = A \mathrm{sin} (k x - \omega t)
The given equation is a variant of this equation. The amplitude is 1A in both.
k x - \omega t = \mathrm{const}
So, by definition, the velocity is in the negative x-direction.
k x = \omega t + \mathrm{const}
x = \frac{\omega t + \mathrm{const}}{k}
\frac{x}{t} = \frac{\omega t + \mathrm{const}}{kt}
v = \frac{\omega}{k} + \mathrm{const}
And for some mysterious
\omega = 2 \pi \lambda
k = 2 \pi T
So plug these in
v = \frac{\omega}{k} = \frac{2 \pi \lambda}{ 2 \pi T } = \frac{\lambda}{T}
The amplitude is A. The velocity for all sine waves are in the positive direction. The answer is E

Problem 11

820

In a double-slit interference experiment, d is the distance between the centers of the slits and w is the width of each slit, as shown in the figure above. For incident plane waves, an interference maximum on a distance screen will be "missing" when

  1. d = √(3) w
  2. 3 d = 2 w
  3. d = √(2) w
  4. 2 d = 3 w
  5. 2 d = w

Solution 11

Answer

Problem 12

Light with wavelength 5200 angstroms is incident normally on a transmission diffraction grating with 2000 lines per centimeter. The first-order diffraction maximum is at an angle with respect to the incident beam, that is most nearly how many degrees?

Solution 12

Diffraction gratings have the same formula as 2-slit interference, except each slit is (obviously) much smaller. The condition for maximum is given by
d\sin\theta = m\lambda
relating the width of the slit to the wavelength and angle and order m. The width of each slit is given by the grating
d=(2000 lines/cm \times 100cm/m)^{-1}=0.5E-5 m
Thus, plugging in the wavelength one has
\sin\theta = \lambda/d = 5200E-10/0.5E-5 \approx 10000E-5 = 1E-1
Now, the approximations to get rid of the trig function. Since
\theta << 1
one can approximate
\sin\theta \approx \theta
where the angle is in radians. Now, convert the angle from radians to degrees.
1E-1 \times 180^{\circ}/\pi = 18/\pi \approx 18/3 = 6^{\circ}

Problem 13

867

A steady beam of light is normally incident on a piece of polaroid. As the polaroid is rotated around the beam axis, the transmitted intesity varies as A + B cos 2θ where θ is the angle of rotation and A and B are constants with A > B > 0. Which of the following maybe be correctly concluded about the incident light?

  1. The light is completely unpolarized
  2. The light is completely plane polarized
  3. The light is completely circularly polarized
  4. The light is partly plane polarized and partly unpolarized
  5. The light is partly circularly polarized and partly unpolarized

Solution 13

Answer

Problem 14

896

A gas-filled cell of length 5 cm is inserted in one arm of a Michelson interferometer, as shown in the figure above. The interferometer is illuminated by light of wavelength 500 nm. As the gas is evacuated from the cell, 40 fringes cross a point in the field of view. What is the refractive index of this gas?

Solution 14

Answer