MP11-2: A 2 lens system (hints - read me first)

Hints Part A


Hints Part B


Hints Part D


Hints Part E

MP11-1: A Microscope for biology (hints)

Hint 2 (d1 = )


Answers for Hints 2 & 3:

MP11-1: A Convex-convex lens (hints - read me first)

Hint #1: Convex-convex tells you whether the lens is converging or diverging. You don't need to know any of the numeric parameters for Part A.

Hint #2: The side that the light leaves the lens, corresponding to R2, is convex, so it is CONCAVE from the light's perspective inside the glass.

Hint #3: R2 is negative.

MP10-2: Reflection and Refraction ranking task (hints)

Hint #1: This is a silly problem.
Hint #2: More than 1 card can go in a 'slot'.

MP9: Problem 22.36 (solution with explanation)

Recall that the intensity is proportional to E^2 where E is the electric field. So, if you have only one half the number of slits, you have one fourth the intensity. 'Nuff said. Peace out.

MP9: Problem 22.18 (solution with explanation)

The key to this problem is the basic diffraction relationship asin (theta) = lambda, where lambda is ( the speed of light 3 * 10^ 8 m/s / the given 800 Mhz Frequency), then you multiply by 180/pi to convert to radians.

However, this gives you only half the angle you need. Recall that for diffraction the 1st minima correspond with m=+/- 1, and the central maximum is everything in between. So you need to multiple the above function by 2.

MP8-2: Lights of different wavelengths on a diffraction grating (hints - read this first)

Use the given slit density of the diffraction grating to determine the value of d (specifically, the value of d is the reciprocal of this slit density, but you must convert to meters first).

MP8-2 Overlapping Diffraction Patterns (hints)

These hints apply to Part C

MP8-1: Fringes from different interfering wavelengths (spoiler with explanation)

Hints for this problem: Start with the constructive interference equation, use the small angle approximation (sin x = x) and solve for theta when m=1 (it isn't m=0 because that would be for theta=0 and we know there is an nonzero angle based on the problem description).

The theta will be the same for the destructive scenario, since the destructive band with the second wavelength occurs in the same place as the constructive band for the original wavelength. Use m=0, since this is the first destructive interference band.

Solving this relationship shows you that the second wavelength is twice the first, which you then convert to micrometers.

MP6-2: Driven LC Circuit (hints only - look at this first)