Calculating Height for Maximum Illumination on a Round Table

Learn how to calculate the height at which to hang a bulb above a round table for maximum illumination. Understand the principles of optics and apply the inverse square law of illuminance and Lambert's cosine law. Use the Pythagorean theorem and differentiate the equation to find the optimal height.

Object – I have a round table which I sometimes use for study purposes. I don’t like putting a
lamp on the table so want to fix a bulb just above the centre of the table. To fulfill the purpose, I
need to calculate the height to which I should hang the bulb so that the maximum illumination
reaches at the edges of the table.
(bulb)
l
θ h
(table)
r
Assumption – As part of physics study I know that some rules of optics are applicable in this
situation:
i. “Inverse Square Law of Illuminance” which states that if light rays incident on an object
obliquely, the illumination is proportional to the square of the distance.
ii. “Lambert's Cosine Law” which states that illumination is directly proportional to the
cosine of the angle of incidence on the surface.
Solution – Total illumination (I) of the lamp is going to be
I ∝ cos θ
I ∝ 1
l2
Total illumination will be
I ∝ cos θ
l2
I =k cos θ
l2 ----------------------------------------------------- equation 1

Here ‘k’ is a constant of proportionality.
Now, we observe the figure we see a right angle triangle
We also know that the Pythagoras theorem is applicable only to the right angle triangle.
According to which-
r2 +h2 =l2
Here –
r = radius of the table
h = height of the bulb
l = oblique distance between the light source and corner of the table
l
h
r
According to the above triangle
cos θ= h
l
cos θ= h
√r 2+ h2
Fromm equation 1, we have
I = k
(r2 +h2 )
h
√r2 +h2
I =k h
(r ¿ ¿2+h2)
3
2 ¿
----------------------------------- equation 2
We know that as we change the height the illumination changes. Now we differentiate equation 2
w.r.t height ‘h’
θ

dI
dh =
k h
(r ¿ ¿2+h2)
3
2 − 3
2
k ( h ) (2 h)
( r ¿¿ 2+ h2)
5
2 ¿
¿ --------------------------------- equation 3
For maxima or minima we put dI
dh = 0
k h
(r ¿ ¿2+ h2)
3
2 − 3
2
k ( h ) ( 2 h )
(r ¿¿ 2+ h2)
5
2 =0 ¿
¿
3 h2=r2+ h2
Finally we get –
h= r
√ 2
now to check the result for maxima or minima we put h= r
√2 in second derivative of equation 1
d2 x
d h2 =
k h
(r ¿ ¿2+ h2)
3
2 −3.2 k h
(r ¿¿ 2+h2)
5
2 − 5
2
h
(r ¿ ¿ 2+ h2)
7
2 ¿
¿
¿ ------------------------------
equation 4
Putting h= r
√ 2 in equation 4, we get
d2 x
d h2 < 0

Thus the height h= r
√2 to the maximum illumination.