1
Quiz 3 solutions are in order of the questions for version A.
(1). Correct answer C
In this situation (R > p > R/2) , the mirror will from a real image with q > p; hence
the image is behind you as you look towards the mirror and you cannot see it.
(2). Correct answer is D
We have to solve
1
2
1
+ =
50 q
60
which yields a positive q (real image) which must therefore be inverted
(3). Correct answer is E
Solving for q in the previous formula gives q = 75cm
(4). Correct answer is A
If we assume that the image is real, we can use q = 60 and find f = 20. We can then
check that this is consistent with the rest of the information given, since this predicts
that q =100 if p =25. The other possibility, that the first image is virtual and has
q=-60 would yield a focal length which disagrees with the second situation.
(5). Correct answer is C
In order to produce a parallel beam to the right, the object for the second lens must
be at p=f, namely 150 cm to the left of the lens. This position is 60cm to the left of
the first lens, which means that the first lens must have made a (virtual) image at q
= -60. We can then determine the focal length via
1
1
1
+
=
72 −60
f1
(6). Correct answer is C
This can be found from the formula
∆y =
Lλ
d
where ∆y is the distance between bright lines.
2
(7). Correct answer is B
We just have
δλ
1
=
λ
Nm
which gives N = 1/3(475/.48) ' 330
(8). Correct answer is A
This is a case where there is an extra phase shift and hence the first constructive interference will happen at a thickness of 1/4 of the wavelength in the soap;
d =.25(660/1.33) giving result A
(9). Correct answer is B
For objects at infinity, the image distance is 50mm; for an object at p=800 mm, the
image distance is easily found to be 53.333; hence the distance must vary by 3.33 mm
(10). Correct answer is B
The angular resolution is 1.44 λ / .85. This must be roughly equal to .03/ L where L
is the distance to the object. This leads to answer B