Cherenkov radiation is emitted by a particle traveling faster than the phase velocity of light in that medium. Because , the phase velocity of light in the glass is:

This is the minimum velocity of a particle to be emitting Cherenkov radiation in glass. This eliminates all answers except  and . The latter is impossible because it is greater than the speed of light in a vacuum, therefore the answer is .

The phase velocity of light through a medium with refractive index  is given by:

Solving this for  and substituting our given value of _:

The condition for constructive interference with double slit diffraction is given by:

Where  is 0, 1, 2, ...

Solving for the angle and using the small angle approximation, we get;

The distance  in the diagram can be related to the other quantities by simple geometry:

Again, with the help of the small angle approximation. Setting the two thetas equal to each other and solving for , we get:

For the central band, , so the  position is also zero. The next band, , yields a distance of:

The beat from two combined sound waves is:

The equation describing maxima of a diffraction grating is:

Where d is the separation of slits, which can also be expressed as:

Substituting d into the first equation and making the small angle approximation, one can solve for  (lines per length):

Note that the angle had to be converted from degrees to radians. Finally, the question asks for lines per centimeter, not meter, so the answer becomes .

The equation for the Doppler effect with frequency for a moving source is given by:

Where  is the speed of sound,  is the speed of the source,  is the stationary frequency and  is the Doppler frequency. Substituting the known values and solving gives us the speed of the car.

Initially unpolarized light passing though a linear polarizer decreases in intensity by a factor of two:

Malus's Law gives the change in intensity of polarized light passing through a linear polarizer in terms of the change in angle of polarization:

Combining the two equations, we get:

Solving for :

Total internal reflection occurs at the angle:

However, this angle is measured from a line normal to the plane of the interface; the angle we want, therefore, is , which is .

Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between  on the right side of the lens.

Alternatively, one can apply the thin lens equation:

Where  is the object distance  and  is the focal length . Plug in these values and solve.

First, find the image distance  from the thin lens equation:

Magnification of a lens is given by:

Where  and  are the image height and object height, respectively. The given object height is , which we can use to solve for the image height:

Because the sign is negative, the image is inverted.