Optics Practice

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1.1Choose the Right Optics Model

Exam I | Problem 1.1 |

A student is studying the bright and dark bands produced by light passing through two slits.

Which optics model should they use?

Solution

The pattern comes from interference and diffraction, so the correct model is wave optics.

Geometric optics treats light as rays, but it does not explain the fringe pattern.

1.2Find a Wavelength in Vacuum

Exam I | Problem 1.2 |

Light in vacuum has frequency

f = 5.0 \times 10^{14}\ \text{Hz}.

What is its wavelength?

Solution

Use

c = \lambda f.

\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7}\ \text{m}.

That is

600\ \text{nm}.

1.3Find Light Speed in Glass

Exam I | Problem 1.3 |

A piece of glass has refractive index

$$ n = 1.5. $$

What is the speed of light in the glass?

Solution

Use

v = \frac{c}{n}.

v = \frac{3.0 \times 10^8}{1.5} = 2.0 \times 10^8\ \text{m/s}.

1.4Refraction Direction at an Interface

Exam I | Problem 1.4 |

Snell's Law, Refraction

A ray of light goes from air into water.

Does it bend toward the normal or away from the normal?

Solution

Water has a larger refractive index than air, so the light slows down as it enters the water.

That means the ray bends toward the normal.

1.5Plane Mirror Image Location

Exam I | Problem 1.5 |

An object is placed 1.8 m in front of a plane mirror.

How far behind the mirror is the image?

Solution

A plane mirror forms an image the same distance behind the mirror as the object is in front of it.

So the image is

1.8\ \text{m}

behind the mirror.

1.6Find a Mirror Focal Length

Exam I | Problem 1.6 |

A concave mirror has radius of curvature

R = 40\ \text{cm}.

What is its focal length?

Solution

For a spherical mirror,

f = \frac{R}{2}.

f = \frac{40}{2} = 20\ \text{cm}.

1.7Solve a Thin Lens Image Distance

Exam I | Problem 1.7 |

Thin Lens Equation

A converging lens has focal length

f = 12\ \text{cm}

and an object is placed

d_o = 36\ \text{cm}

in front of the lens.

Find the image distance.

Solution

Use the thin lens equation:

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}.

Substitute the values:

\frac{1}{12} = \frac{1}{36} + \frac{1}{d_i}.

Subtract:

\frac{1}{d_i} = \frac{1}{12} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18}.

d_i = 18\ \text{cm}.

1.8Find the Magnification

Exam I | Problem 1.8 |

In the previous lens setup, what is the magnification?

Solution

Use

m = -\frac{d_i}{d_o}.

Substitute:

m = -\frac{18}{36} = -\frac{1}{2}.

The negative sign means the image is inverted, and the magnitude shows it is half the object size.

1.9Compute a Critical Angle

Exam I | Problem 1.9 |

Light goes from a material with index

$$ n_1 = 2.0 $$

into a material with index

$$ n_2 = 1.0. $$

What is the critical angle?

Solution

Use

\sin \theta_c = \frac{n_2}{n_1}.

\sin \theta_c = \frac{1.0}{2.0} = \frac{1}{2}.

Therefore,

\theta_c = 30^\circ.

1.10Intensity Through a Polarizer

Exam I | Problem 1.10 |

Linearly polarized light with intensity

$$ I_0 $$

passes through an ideal polarizer whose axis makes a 60 degree angle with the light's polarization direction.

What is the transmitted intensity?

Solution

Use Malus's law:

I = I_0 \cos^2 \theta.

With $\theta = 60^\circ$,

I = I_0 \cos^2 60^\circ = I_0 \left(\frac{1}{2}\right)^2 = \frac{I_0}{4}.

Difficulty

Integrated

2.1A Converging Lens with Object Inside the Focal Length

Exam II | Problem 2.1 |

A converging lens has focal length

f = 15\ \text{cm}.

An object is placed

d_o = 10\ \text{cm}

in front of the lens.

Find the image distance, and state whether the image is real or virtual.

Solution

Use the thin lens equation:

\frac{1}{15} = \frac{1}{10} + \frac{1}{d_i}.

\frac{1}{d_i} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30}.

Thus

d_i = -30\ \text{cm}.

The negative sign means the image is virtual and on the same side as the object.

The magnification is

m = -\frac{d_i}{d_o} = -\frac{-30}{10} = 3.

So the image is upright and enlarged.

2.2Refraction and Speed Change in Water

Exam II | Problem 2.2 |

Snell's Law, Refractive Index, Wave Speed

Light travels from air into water. The angle of incidence in air is

30^\circ

and the refractive index of water is

$$ n = 1.33. $$

Find the refracted angle in water and the speed of light in water.

Solution

Use Snell's law with $n_1 \approx 1$ for air:

n_1 \sin \theta_1 = n_2 \sin \theta_2.

\sin \theta_2 = \frac{1}{1.33}\sin 30^\circ = \frac{1}{1.33}\cdot \frac{1}{2} \approx 0.376.

That gives

\theta_2 \approx 22^\circ.

The speed in water is

v = \frac{c}{n} = \frac{3.0 \times 10^8}{1.33} \approx 2.26 \times 10^8\ \text{m/s}.

2.3Two Lenses in Series

Exam II | Problem 2.3 |

A converging lens with focal length

f_1 = 10\ \text{cm}

forms an image from an object placed

d_{o1} = 30\ \text{cm}

in front of it.

A second converging lens with focal length

f_2 = 15\ \text{cm}

is placed 20 cm to the right of the first lens.

Find the final image position relative to the second lens and describe the final image.

Solution

First lens:

\frac{1}{10} = \frac{1}{30} + \frac{1}{d_{i1}}

\frac{1}{d_{i1}} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15},

and

d_{i1} = 15\ \text{cm}.

That image is 15 cm to the right of the first lens, so it is 5 cm to the left of the second lens. Thus the second lens has object distance

d_{o2} = 5\ \text{cm}.

Second lens:

\frac{1}{15} = \frac{1}{5} + \frac{1}{d_{i2}}.

\frac{1}{d_{i2}} = \frac{1}{15} - \frac{1}{5} = -\frac{2}{15},

and

d_{i2} = -7.5\ \text{cm}.

The negative sign means the final image is virtual and on the left side of the second lens.

The magnifications are

m_1 = -\frac{15}{30} = -\frac{1}{2}, \qquad m_2 = -\frac{-7.5}{5} = \frac{3}{2}.

So the total magnification is

m = m_1 m_2 = -\frac{3}{4}.

The final image is virtual, inverted relative to the original object, and slightly smaller.

2.4Correct a Nearsighted Eye

Exam II | Problem 2.4 |

A student can see nearby objects clearly, but distant street signs are blurry because the image would form in front of the retina.

What vision defect is this, and what kind of corrective lens is used?

Solution

This is myopia or nearsightedness.

It is corrected with a diverging lens, which spreads incoming rays slightly so the eye focuses the image farther back, onto the retina.

2.5Bright or Dark in a Thin Film?

Exam II | Problem 2.5 |

A thin film produces a path difference equal to

\frac{\lambda}{2}

between two reflected rays. Also, one of the reflections occurs from a lower-index medium to a higher-index medium, so that reflection adds a phase shift of $\pi$.

Is the reflected light bright or dark?

Solution

The path difference of $\lambda/2$ contributes a phase shift of $\pi$.

The reflection from lower index to higher index adds another phase shift of $\pi$.

The total phase difference is therefore

\pi + \pi = 2\pi,

which corresponds to constructive interference.

So the reflected light is bright.

2.6Angle of a Diffraction Grating Maximum

Exam II | Problem 2.6 |

A diffraction grating has slit spacing

d = 2.0 \times 10^{-6}\ \text{m}.

Light of wavelength

\lambda = 500\ \text{nm}

falls on the grating.

What is the angle of the second-order bright maximum?

Solution

Use the grating equation:

d \sin \theta = m\lambda.

For second order, $m=2$:

2.0 \times 10^{-6} \sin \theta = 2(500 \times 10^{-9}).

\sin \theta = \frac{1.0 \times 10^{-6}}{2.0 \times 10^{-6}} = 0.5.

Therefore,

\theta = 30^\circ.

2.7Two Polarizers in Sequence

Exam II | Problem 2.7 |

Unpolarized light with intensity

$$ I_0 $$

passes through an ideal polarizer, then a second polarizer set at 30 degrees relative to the first.

What is the final intensity?

Solution

Unpolarized light loses half its intensity in the first polarizer:

I = \frac{I_0}{2}.

Through the second polarizer, Malus's law gives

I = \frac{I_0}{2}\cos^2 30^\circ.

Since

\cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4},

the final intensity is

\frac{I_0}{2}\cdot \frac{3}{4} = \frac{3I_0}{8}.

2.8Choose the Better Resolving Aperture

Exam II | Problem 2.8 |

Two telescopes observe the same light. Telescope A has aperture diameter 5.0 cm, and telescope B has aperture diameter 10.0 cm.

Which telescope has the smaller minimum resolvable angle, and by what factor?

Solution

The Rayleigh criterion says

\theta_{\min} \approx 1.22 \frac{\lambda}{D}.

For the same wavelength, $\theta_{\min}$ is inversely proportional to aperture diameter.

Since telescope B has twice the diameter of telescope A, its minimum resolvable angle is half as large.

So telescope B has the better resolution, by a factor of 2.

Difficulty

Applied

3.1Choose the Correct Vision Aid

Final | Problem 3.1 |

A person can read a book clearly, but distant road signs are blurry.

What type of corrective lens should they use, and why?

Solution

This is myopia.

A diverging lens is used because it reduces the convergence of incoming parallel rays so the eye can focus the image onto the retina instead of in front of it.

3.2Check Total Internal Reflection in a Fiber

Final | Problem 3.2 |

A fiber-optic core has refractive index

$$ n_1 = 1.50, $$

and the cladding has refractive index

$$ n_2 = 1.30. $$

A ray inside the core hits the boundary at an angle of 65 degrees from the normal.

Will the ray undergo total internal reflection?

Solution

First find the critical angle:

\sin \theta_c = \frac{n_2}{n_1} = \frac{1.30}{1.50} \approx 0.867.

\theta_c \approx 60^\circ.

The incident angle is 65 degrees, which is larger than the critical angle.

Therefore the ray undergoes total internal reflection and stays confined in the fiber.

3.3Which Instrument Fits the Job?

Final | Problem 3.3 |

You want a distant planet to appear larger in angular size.

Should you use a microscope or a telescope?

Solution

Use a telescope.

Microscopes are for very small nearby objects, while telescopes are designed to view far objects with larger angular size.

3.4Compare Two Apertures

Final | Problem 3.4 |

Two cameras use the same wavelength of light.

Camera A has an aperture diameter of 4.0 cm, and camera B has an aperture diameter of 10.0 cm.

Which camera has the better resolution?

Solution

By the Rayleigh criterion,

\theta_{\min} \approx 1.22 \frac{\lambda}{D}.

For the same wavelength, a larger aperture means a smaller minimum resolvable angle.

Since camera B has the larger aperture, it has the better resolution.

In fact, its minimum angle is smaller by a factor of

\frac{10.0}{4.0} = 2.5.

3.5Which Color Bends More in a Prism?

Final | Problem 3.5 |

White light passes through a prism.

Which color bends more in normal dispersion, red or blue?

Solution

In normal dispersion, shorter wavelengths have larger refractive index.

Blue light has a shorter wavelength than red light, so it bends more.

Therefore, blue bends more than red.

Difficulty

Challenge

4.1Two-Lens System With a Final Virtual Image

Final | Problem 4.1 |

A converging lens has focal length

f_1 = 12\ \text{cm}

and forms an image from an object placed

d_{o1} = 18\ \text{cm}

in front of it.

A diverging lens with focal length

f_2 = -28\ \text{cm}

is placed 50 cm to the right of the first lens.

Find the final image position relative to the second lens, and state whether the final image is upright or inverted.

Solution

For the first lens:

\frac{1}{12} = \frac{1}{18} + \frac{1}{d_{i1}}.

\frac{1}{d_{i1}} = \frac{1}{12} - \frac{1}{18} = \frac{1}{36},

and

d_{i1} = 36\ \text{cm}.

That image is 36 cm to the right of the first lens, so it is 14 cm to the left of the second lens. Thus the second lens has object distance

d_{o2} = 14\ \text{cm}.

For the second lens:

\frac{1}{-28} = \frac{1}{14} + \frac{1}{d_{i2}}.

\frac{1}{d_{i2}} = -\frac{1}{28} - \frac{1}{14} = -\frac{3}{28},

which gives

d_{i2} = -\frac{28}{3}\ \text{cm}.

The negative sign means the final image is virtual and on the left side of the second lens.

The magnifications are

m_1 = -\frac{36}{18} = -2, \qquad m_2 = -\frac{-28/3}{14} = \frac{2}{3}.

So the total magnification is

m = m_1 m_2 = -\frac{4}{3}.

The final image is virtual, inverted, and enlarged.

4.2Constructive Thin-Film Reflection

Final | Problem 4.2 |

A thin film has refractive index

$$ n = 1.50 $$

and is surrounded by air. Light of wavelength

\lambda = 600\ \text{nm}

in air strikes the film.

What is the smallest film thickness that gives constructive reflection?

Solution

Because the reflection from air to film is from lower index to higher index, one reflected ray gets a phase shift of $\pi$.

For constructive reflection with one phase reversal, the film thickness must satisfy

2nt = \left(m + \frac{1}{2}\right)\lambda.

The smallest positive thickness uses $m=0$:

2(1.50)t = \frac{1}{2}(600\ \text{nm}).

3t = 300\ \text{nm}

and

t = 100\ \text{nm}.

4.3Compare Two Resolving Powers

Final | Problem 4.3 |

Two telescopes observe light of wavelength

500\ \text{nm}.

Telescope A has aperture diameter 4.0 cm, and telescope B has aperture diameter 10.0 cm.

Find the Rayleigh minimum angle for each telescope, and state which one resolves finer detail.

Solution

Use

\theta_{\min} \approx 1.22 \frac{\lambda}{D}.

For telescope A:

\theta_A = 1.22 \frac{5.0 \times 10^{-7}}{0.040} = 1.525 \times 10^{-5}\ \text{rad}.

For telescope B:

\theta_B = 1.22 \frac{5.0 \times 10^{-7}}{0.10} = 6.10 \times 10^{-6}\ \text{rad}.

Because telescope B has the smaller minimum angle, it resolves finer detail.

Its minimum angle is smaller by a factor of

\frac{1.525 \times 10^{-5}}{6.10 \times 10^{-6}} = 2.5.

4.4Three Polarizers With a Fixed Output

Final | Problem 4.4 |

Unpolarized light of intensity $I_0$ passes through three polarizers.

The first and third polarizers are crossed, and the middle polarizer is rotated by an angle $\theta$ from the first.

If the final intensity is

\frac{I_0}{16},

what is one possible value of $\theta$?

Solution

After the first polarizer, the intensity is

\frac{I_0}{2}.

After the middle polarizer,

I = \frac{I_0}{2}\cos^2 \theta.

The third polarizer is crossed with the first, so it makes an angle of $90^\circ - \theta$ with the middle one. The final intensity is

I = \frac{I_0}{2}\cos^2 \theta \sin^2 \theta.

Use

\sin 2\theta = 2\sin\theta\cos\theta

to rewrite this as

I = \frac{I_0}{8}\sin^2(2\theta).

Set this equal to $I_0/16$:

\frac{I_0}{8}\sin^2(2\theta) = \frac{I_0}{16}.

\sin^2(2\theta) = \frac{1}{2}.

One possible choice is

2\theta = 45^\circ,

which gives

\theta = 22.5^\circ.

Because of symmetry, $67.5^\circ$ is also a valid answer.

Optics Practice

Selected problems

Summary

Direct

1.1Choose the Right Optics Model

1.2Find a Wavelength in Vacuum

1.3Find Light Speed in Glass

1.4Refraction Direction at an Interface

1.5Plane Mirror Image Location

1.6Find a Mirror Focal Length

1.7Solve a Thin Lens Image Distance

1.8Find the Magnification

1.9Compute a Critical Angle

1.10Intensity Through a Polarizer

Integrated

2.1A Converging Lens with Object Inside the Focal Length

2.2Refraction and Speed Change in Water

2.3Two Lenses in Series

2.4Correct a Nearsighted Eye

2.5Bright or Dark in a Thin Film?

2.6Angle of a Diffraction Grating Maximum

2.7Two Polarizers in Sequence

2.8Choose the Better Resolving Aperture

Applied

3.1Choose the Correct Vision Aid

3.2Check Total Internal Reflection in a Fiber

3.3Which Instrument Fits the Job?

3.4Compare Two Apertures

3.5Which Color Bends More in a Prism?

Challenge

4.1Two-Lens System With a Final Virtual Image

4.2Constructive Thin-Film Reflection

4.3Compare Two Resolving Powers

4.4Three Polarizers With a Fixed Output