Two Lamps

Consider two scenarios, one in which the dim lamp is close (Case 1), and the other in which the dim lamp is far (Case 2). By assumption, $I>i$ .

Let us consider the distances $D > d > 0$ . From the inverse square law, we know that the amount of light recieved by the desk is of the form $L=k \frac{i}{r^2}$ , where $L$ is the amount of light, $k$ is some positive proportionality constant, $i$ is the intensity of the source and $r$ is the distance from the source.

Therefore, the amount of light the desk recieved in the first case is $L_1=k \frac{i}{d^2}+k \frac{I}{D^2}=k({\frac{i}{d^2}+\frac{I}{D^2}})$ . Similarly, the light the desk recieved in the second case is is $L_2=k({\frac{I}{d^2}+\frac{i}{D^2}})$ .

If we assume that $L_1\geq L_2$ , then

$k({\frac{I}{d^2}+\frac{i}{D^2}})\leq k({\frac{i}{d^2}+\frac{I}{D^2}})$

$\Rightarrow {\frac{I}{d^2}+\frac{i}{D^2}}\leq{\frac{i}{d^2}+\frac{I}{D^2}}$

Multiplying both sides by $D^2 d^2$ yields

$\Rightarrow ID^2+id^2\leq Id^2+iD^2$

$\Rightarrow I(D^2-d^2) \leq i(D^2-d^2)$

$\therefore I\leq i$

which contradicts our assumption that $I>i$ . Therefore it is the case that $L_2>L_1$ . In other words, the dimmer lamp is farther away, and the lamp that has been replaced is the one closer to the desk.

your desk would have been less lit if the other lamp happened to be the one substituted = your desk would have been more lit if the lamp happened to be the one substituted, that must be the lamp close to the desk.

great question, made me think

"your desk would have been less lit if the other lamp happened to be the one substituted" => the one substituted is the one with predominant lighting of your desk = the closest one. The "inverse square" hint is meant to create confusion. It's a logic problem, not physics.

The illumination from the original bulb (b) closer to the desk is (D/d)^2 brighter than that from the farther identical bulb. D>d.

Now we place a brighter bulb (illumination B > b) in one location. Again, if we place it closer the illumination on the desk will increase (D/d)^2 times than if we place it farther.

We lose the original illumination b, so the final illumination is B(D/d)^2 + b if we place it closer and b(D/d)^2 + B if we place it farther.

D/d is > 1 because D>d.

Thus (D/d)^2 >1

Let (D/d)^2 = n

Thus n > 1

Bn + b > bn + B when n > 1 and B > b (I.e. total illumination on the desk is greater when the brighter bulb is placed closer).

Proof: subtract B from each side

B(n-1) + b > bn

Subtract b from each side

B(n-1) > b(n-1)

Divide each side by (n-1). This is a positive number because we know n > 1

B > b which the problem states.

QED