Although it is possible in principle to have an object so far away that it is practically at infinity,
that is not possible in this situation, since Zahra can’t take her lens very far away from the
fire. By the way, this means that thefocal lengthfis not where thefocushappens — the focus
happens atdi.
For similar reasons, we can’t havedo=f.
Since all the variables are positive, we must have 1/doand 1/diboth less than 1/f. This
implies thatdo> fanddi> f. Of the nine logical possibilities in the table, only this one is
actually possible for this real image.
Solutions for chapter 13
Page 950, problem 48:
The expressions|Ψ|^2 and|Ψ^2 |are identical, because the magnitude of a product is the product
of the magnitudes. These expressions give positive real numbers as their results, which makes
sense for a probability density. The expression Ψ^2 need not be real, and if it is real, it may be
negative. It cannot be interpreted as a probability density. As a concrete example, suppose that
Ψ =bi, wherebis a real number with units. Then|Ψ|^2 =|Ψ^2 |=b^2 , which is real and positive,
but Ψ^2 =−b^2 , which clearly can’t be interpreted probabilistically, because it’s negative.
Page 950, problem 49:
(a) The quantityx−yvanishes along the liney=xlying in the first quadrant at a 45-degree
angle between the axes. Squaring produces a trough parallel to this line, with a parabolic cross-
section. Geometrically, the Laplacian can be interpreted as a measure of how much the value of
fat a point differs from its average value on a small circle centered on that point. The trough
is concave up, so we can predict that the Laplacian will be positive everywhere.
(b) The zero result is clearly wrong because it disagrees with our conclusion from part a that
the Laplacian is positive. A correct calculation gives∂^2 (x−y)^2 /∂x^2 +∂^2 (x−y)^2 /∂y^2 = 4.
(c) If we rotate our coordinate axes counterclockwise by 45 degrees, then we have a parabolic
trough oriented along thexaxis. In terms of these new coordinates,∂f/∂x= 0, while∂f/∂yis
nonzero almost everywhere.
Remark: The mistake described in the question is a common one, and is apparently based on
the idea that the notation∇^2 must mean applying an operator∇twice. For those with some
exposure to vector calculus, it may be of interest to note that the Laplacianisequivalent to the
divergence of the gradient, which can be notated either div(gradf) or∇·(∇f). The important
thing to recognize is that the gradient, notated gradfor∇f, outputs avector, not a scalar like
the quantityQdefined in this problem.
Solutions for chapter 14
Page 1008, problem 2:
{xˆ}is not a basis, because there are vectors such asyˆthat we can’t form as a linear combination
(i.e., scalar multiple) ofˆx. {xˆ,ˆy}is the standard basis for this vector space. {−ˆx,ˆx+yˆ}also
works as a basis, because the two vectors are linearly independent, and it’s easy to check that
any vector in the plane can be formed as a linear superposition of them.{xˆ,ˆy,xˆ+yˆ}is not a
basis, because these three vectors are not linearly independent.
Page 1008, problem 3:
(a) The sketch forwill be a 45-degree line through the origin, whilerwill be only the part of that line in the first quadrant. Of the two, onlyis a vector space. The setrisn’t a vector
space, because it doesn’t have additive inverses.