Optics Practice

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.

Use

So

That is

Use

So

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.

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

So the image is

behind the mirror.

For a spherical mirror,

So

Use the thin lens equation:

Substitute the values:

Subtract:

So

Use

Substitute:

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

Use

So

Therefore,

Use Malus's law:

With $\theta = 60^\circ$,

Use the thin lens equation:

So

Thus

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

The magnification is

So the image is upright and enlarged.

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

So

That gives

The speed in water is

First lens:

so

and

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

Second lens:

So

and

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

The magnifications are

So the total magnification is

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

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.

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

which corresponds to constructive interference.

So the reflected light is bright.

Use the grating equation:

For second order, $m=2$:

So

Therefore,

Unpolarized light loses half its intensity in the first polarizer:

Through the second polarizer, Malus's law gives

Since

the final intensity is

The Rayleigh criterion says

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.

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.

First find the critical angle:

So

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.

Use a telescope.

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

By the Rayleigh criterion,

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

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.

For the first lens:

So

and

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

For the second lens:

So

which gives

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

The magnifications are

So the total magnification is

The final image is virtual, inverted, and enlarged.

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

The smallest positive thickness uses $m=0$:

So

and

Use

For telescope A:

For telescope B:

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

Its minimum angle is smaller by a factor of

After the first polarizer, the intensity is

After the middle polarizer,

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

Use

to rewrite this as

Set this equal to $I_0/16$:

So

One possible choice is

which gives

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