LAWS OF

ILLUMINATION
E E 5 4 7 U T I L I Z AT I O N O F E L E C T R I C A L E N E R GY E N G . YA S M I N M O H A M E D
Inverse square law :
𝐴1
→ 𝑓𝑜𝑟 𝐴1 ∶ ω = 2
𝑑
𝐹 𝐴1
𝐼= →𝐹 =𝐼∗ω= 𝐼∗ 2
ω 𝑑
𝐹
𝐸=
𝐴
𝐴1 1 𝐼
𝐸1 = 𝐼 ∗ 2 ∗ = 2
𝑑 𝐴1 𝑑
𝐼
𝐸2 =
(2𝑑)2
𝐼 𝐼 𝐼
𝐸1 ∶ 𝐸2 ∶ 𝐸3 = 2 ∶ ∶
𝑑 (2𝑑)2 (3𝑑)2
Lambert's cosine law :
𝐼 𝐼 ℎ ℎ
𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ , cos θ = →𝑑=
𝑑 ℎ 𝑑 cos θ
Sheet #1:
4) The candle power of a lamp is 120. A plane surface is placed at a distance of 2.5 m from this lamp.
Calculate the illumination on the surface when it is:
i- normal to rays. 𝐴𝑟𝑐
→ 𝑃𝑙𝑎𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 ∶ 𝜃 = (𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
ii- inclined to the ray by 45°. 𝑟
𝐴 𝜃
iii- parallel to the rays. → 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑔𝑙𝑒 ∶ ω = 2 = 2𝜋 1 − cos (𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝑟 2
𝐹
𝐼 = 120 𝐶𝑃 , 𝑑 = 2.5 𝑚 → 𝐿𝑢𝑚𝑖𝑛𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∶ 𝐼 = (𝐿𝑢𝑚𝑒𝑛Τ𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 𝑜𝑟 (𝑐𝑑)
ω
𝐼𝑢𝑚𝑒𝑛𝑠(𝐹)
→ 𝐶𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝐶. 𝑃 = (𝑐𝑑)
ω
𝐹
→ 𝐼𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (luminance): 𝐸 = (𝐿𝑢𝑚𝑒𝑛Τ𝑚2 ) 𝑜𝑟 (𝑙𝑢𝑥)
𝐼 120 𝐴
𝐼
𝑖) 𝐸 = 2 = 2
= 19.2 𝑙𝑢𝑥 → 𝐵𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠: 𝐿 = 𝑐𝑑 Τ𝑚2
𝑑 2.5 𝐴 cos 𝜃
⇒ 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑒 𝑠ℎ𝑝ℎ𝑒𝑟𝑎 ∶ 𝐸 = 𝜋 𝐿
𝐹
𝐼 120 → 𝑀𝑒𝑎𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝑀𝑆𝐶𝑃 =
4𝜋
(𝑐𝑑)
𝑖𝑖) 𝐸 = 2 cos θ = cos 45 = 13.58 𝑙𝑢𝑥 𝐹
𝑑 2.52 → 𝐿𝑎𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 ∶ η =
𝑃
→ 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑟𝑒𝑎𝑐ℎ𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑛
𝑖𝑖𝑖) 𝐸 = 0 𝑈𝐹 =
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑒𝑚𝑖𝑡𝑡𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒
→ 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸) 𝑢𝑛𝑑𝑒𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑀𝐹 =
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸)𝑢𝑛𝑑𝑒𝑟 𝑒𝑣𝑒𝑟𝑦 𝑡ℎ𝑖𝑛𝑔 𝑖𝑠 𝑐𝑙𝑒𝑎𝑛
1
→ 𝐷𝑒𝑝𝑒𝑟𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶ 𝐷𝐹 =
𝑀𝐹
𝐼 𝐼
→ 𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ
𝑑 ℎ
5) A 500 W lamp having MSCP of 800 is suspended 3 m above the working plane. Find:
i- illumination directly below the lamp at the working plane.
ii- lamp efficiency.
iii- illumination at a point on the horizontal plane 2.4 m away from vertically below the lamp.
𝐴𝑟𝑐
→ 𝑃𝑙𝑎𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 ∶ 𝜃 = (𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝑃 = 500 𝑊 , 𝑀𝑆𝐶𝑃 = 800 𝑐𝑑 , ℎ = 3 𝑚 𝑟
𝐴 𝜃
→ 𝐸 directly below the lamp =? ? ? → η =? ? ? → 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑔𝑙𝑒 ∶ ω = 2 = 2𝜋 1 − cos
𝑟 2
(𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
→ 𝐸 at a point on the horizontal plane 2.4 m =? ? ? → 𝐿𝑢𝑚𝑖𝑛𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∶ 𝐼 =
𝐹
(𝐿𝑢𝑚𝑒𝑛Τ𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 𝑜𝑟 (𝑐𝑑)
ω
𝐼𝑢𝑚𝑒𝑛𝑠(𝐹)
𝐼 800 → 𝐶𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝐶. 𝑃 =
ω
(𝑐𝑑)
𝑖) 𝐸 = 2 = 2 = 88.9 𝑙𝑢𝑥 𝐹
ℎ 3 → 𝐼𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (luminance): 𝐸 = (𝐿𝑢𝑚𝑒𝑛Τ𝑚2 ) 𝑜𝑟 (𝑙𝑢𝑥)
𝐴
𝐹 𝐼
→ 𝐵𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠: 𝐿 = 𝑐𝑑 Τ𝑚2
𝑖𝑖) η = 𝐴 cos 𝜃
𝑃 ⇒ 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑒 𝑠ℎ𝑝ℎ𝑒𝑟𝑎 ∶ 𝐸 = 𝜋 𝐿
𝐹 → 𝑀𝑒𝑎𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝑀𝑆𝐶𝑃 =
𝐹
(𝑐𝑑)
𝑀𝑆𝐶𝑃 = → 𝐹 = 𝑀𝑆𝐶𝑃 ∗ 4𝜋 4𝜋
4𝜋 𝐹
→ 𝐿𝑎𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 ∶ η =
𝑃
𝐹 = 800 ∗ 4𝜋 = 3200 𝜋 𝑙𝑢𝑚𝑒𝑛𝑠 → 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑟𝑒𝑎𝑐ℎ𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑛
𝑈𝐹 =
𝐹 3200 𝜋 𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑒𝑚𝑖𝑡𝑡𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒
η= = = 20.1 𝑙𝑢𝑚𝑒𝑛𝑠/𝑤𝑎𝑡𝑡 → 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑃 500 𝑀𝐹 =
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸) 𝑢𝑛𝑑𝑒𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸)𝑢𝑛𝑑𝑒𝑟 𝑒𝑣𝑒𝑟𝑦 𝑡ℎ𝑖𝑛𝑔 𝑖𝑠 𝑐𝑙𝑒𝑎𝑛
3 1
𝐼 3
800 3 → 𝐷𝑒𝑝𝑒𝑟𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶ 𝐷𝐹 =
𝑖𝑖𝑖) 𝐸 = 2 𝑐𝑜𝑠 θ = 2 = 42.3 𝑙𝑢𝑥 𝑀𝐹
𝐼 𝐼
ℎ 3 32 + 2.42 → 𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ
𝑑 ℎ
6) A lamp is mounted 10 m above the center of a circular area of 20 m diameter. If the lamp gives a uniform CP of 800 over the
circular area, determine the maximum and minimum illumination produced on the area.

20 𝐴𝑟𝑐
ℎ = 10 𝑚 , 𝑟 = = 10 𝑚 , 𝐼 = 800 𝐶𝑃 → 𝑃𝑙𝑎𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 ∶ 𝜃 = (𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
2 𝑟
→ 𝐸𝑚𝑎𝑥 =? ? ? → 𝐸𝑚𝑖𝑛 =? ? ? 𝐴 𝜃
→ 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑔𝑙𝑒 ∶ ω = 2 = 2𝜋 1 − cos (𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝑟 2
𝐹
→ 𝐿𝑢𝑚𝑖𝑛𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∶ 𝐼 = (𝐿𝑢𝑚𝑒𝑛Τ𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 𝑜𝑟 (𝑐𝑑)
ω
𝐼𝑢𝑚𝑒𝑛𝑠(𝐹)
→ 𝐶𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝐶. 𝑃 = (𝑐𝑑)
ω
𝐹
→ 𝐼𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (luminance): 𝐸 = (𝐿𝑢𝑚𝑒𝑛Τ𝑚2 ) 𝑜𝑟 (𝑙𝑢𝑥)
𝐴
𝐼
→ 𝐵𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠: 𝐿 = 𝑐𝑑 Τ𝑚2
𝐴 cos 𝜃
⇒ 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑒 𝑠ℎ𝑝ℎ𝑒𝑟𝑎 ∶ 𝐸 = 𝜋 𝐿
𝐼 800 𝐹
𝐸𝑚𝑎𝑥 = 2 = 2 = 8 𝑙𝑢𝑥 → 𝑀𝑒𝑎𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝑀𝑆𝐶𝑃 = (𝑐𝑑)
ℎ 10 4𝜋
𝐹
→ 𝐿𝑎𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 ∶ η =
𝑃
3 → 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝐼 3
800 10 𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑟𝑒𝑎𝑐ℎ𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑛
𝐸𝑚𝑖𝑛 = 2 𝑐𝑜𝑠 θ = 2 = 2.83 𝑙𝑢𝑥 𝑈𝐹 =
ℎ 10 102 + 102 𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑒𝑚𝑖𝑡𝑡𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒
→ 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸) 𝑢𝑛𝑑𝑒𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑀𝐹 =
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸)𝑢𝑛𝑑𝑒𝑟 𝑒𝑣𝑒𝑟𝑦 𝑡ℎ𝑖𝑛𝑔 𝑖𝑠 𝑐𝑙𝑒𝑎𝑛
1
→ 𝐷𝑒𝑝𝑒𝑟𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶ 𝐷𝐹 =
𝑀𝐹
𝐼 𝐼
→ 𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ
𝑑 ℎ
7) A lamp having a uniform CP of 300 in all directions is provided with a reflector which directs 60% of the total light uniformly
on to a circular area of 12 m diameter. The lamp is 5 m above the area. Calculate:
i- the illumination at the center and edge of the surface with and without reflector.
ii- the average illumination over the area without the reflector. 𝐴𝑟𝑐
→ 𝑃𝑙𝑎𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 ∶ 𝜃 = (𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝐶𝑃 = 300 𝐶𝑃 , 𝑟𝑒𝑓𝑙𝑐𝑡𝑜𝑟 = 60% , 𝑟 = 12/2 = 6 𝑚 , ℎ = 5 𝑚 𝑟
𝐴 𝜃
→ 𝐸𝑐𝑒𝑛𝑡𝑒𝑟 &𝐸𝑒𝑔𝑑𝑒 =? ? ? 𝑤𝑖𝑡ℎ 𝑎𝑛𝑑 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 → 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑔𝑙𝑒 ∶ ω = 2 = 2𝜋 1 − cos (𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝑟 2
→ 𝐸𝑎𝑣 =? ? ? 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 𝐹
→ 𝐿𝑢𝑚𝑖𝑛𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∶ 𝐼 = (𝐿𝑢𝑚𝑒𝑛Τ𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 𝑜𝑟 (𝑐𝑑)
ω
𝐼𝑢𝑚𝑒𝑛𝑠(𝐹)
→ 𝐶𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝐶. 𝑃 = (𝑐𝑑)
ω
𝐹
→ 𝐼𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (luminance): 𝐸 = (𝐿𝑢𝑚𝑒𝑛Τ𝑚2 ) 𝑜𝑟 (𝑙𝑢𝑥)
𝒊) 𝒘𝒊𝒕𝒉𝒐𝒖𝒕 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒐𝒓 ∶ 𝐴
𝐼
→ 𝐵𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠: 𝐿 = 𝑐𝑑 Τ𝑚2
𝐼 300 𝐴 cos 𝜃
⇒ 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑒 𝑠ℎ𝑝ℎ𝑒𝑟𝑎 ∶ 𝐸 = 𝜋 𝐿
𝐸𝑐𝑒𝑛𝑡𝑒𝑟 = 2 = 2 = 12 𝑙𝑢𝑥 𝐹
ℎ 5 → 𝑀𝑒𝑎𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝑀𝑆𝐶𝑃 = (𝑐𝑑)
3 4𝜋
𝐼 300 5 𝐹
3 → 𝐿𝑎𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 ∶ η =
𝐸𝑒𝑑𝑔𝑒 = 2 𝑐𝑜𝑠 θ = 2 𝑃
ℎ 5 52 + 62 → 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑟𝑒𝑎𝑐ℎ𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑛
𝑈𝐹 =
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑒𝑚𝑖𝑡𝑡𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒
= 3.15 𝑙𝑢𝑥 → 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸) 𝑢𝑛𝑑𝑒𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑀𝐹 =
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸)𝑢𝑛𝑑𝑒𝑟 𝑒𝑣𝑒𝑟𝑦 𝑡ℎ𝑖𝑛𝑔 𝑖𝑠 𝑐𝑙𝑒𝑎𝑛
1
→ 𝐷𝑒𝑝𝑒𝑟𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶ 𝐷𝐹 =
𝑀𝐹
𝐼 𝐼
→ 𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ
𝑑 ℎ
 𝐶𝑃 = 300 𝐶𝑃 , 𝑟𝑒𝑓𝑙𝑐𝑡𝑜𝑟 = 60% , 𝑟 = 12/2 = 6 𝑚 , ℎ = 5 𝑚 𝐴𝑟𝑐
→ 𝑃𝑙𝑎𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 ∶ 𝜃 = (𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝑟
→ 𝐸𝑐𝑒𝑛𝑡𝑒𝑟 &𝐸𝑒𝑔𝑑𝑒 =? ? ? 𝑤𝑖𝑡ℎ 𝑎𝑛𝑑 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 𝐴 𝜃
→ 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑔𝑙𝑒 ∶ ω = 2 = 2𝜋 1 − cos (𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
→ 𝐸𝑎𝑣 =? ? ? 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 𝑟 2
𝐹
→ 𝐿𝑢𝑚𝑖𝑛𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∶ 𝐼 = (𝐿𝑢𝑚𝑒𝑛Τ𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 𝑜𝑟 (𝑐𝑑)
𝒊𝒊)𝑬𝒂𝒗 𝒘𝒊𝒕𝒉𝒐𝒖𝒕 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒐𝒓 ∶ ω
𝐼𝑢𝑚𝑒𝑛𝑠(𝐹)
𝐹 → 𝐶𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝐶. 𝑃 =
ω
(𝑐𝑑)
𝐸𝑎𝑣 = &𝐹 =𝐼∗ω 𝐹
𝐴 → 𝐼𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (luminance): 𝐸 = (𝐿𝑢𝑚𝑒𝑛Τ𝑚2 ) 𝑜𝑟 (𝑙𝑢𝑥)
𝛼 5 𝐴
𝐼
ω = 2𝜋 1 − cos = 2𝜋 1 − cos 𝜃 = 2𝜋 1 − → 𝐵𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠: 𝐿 = 𝑐𝑑 Τ𝑚2
2 52 + 62 𝐴 cos 𝜃
⇒ 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑒 𝑠ℎ𝑝ℎ𝑒𝑟𝑎 ∶ 𝐸 = 𝜋 𝐿
= 1.28𝜋 𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠 → 𝑀𝑒𝑎𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝑀𝑆𝐶𝑃 =
𝐹
(𝑐𝑑)
4𝜋
𝐹 = 𝐼 ∗ ω = 300 ∗ 1.28𝜋 = 384 𝜋 𝑙𝑢𝑚𝑒𝑛𝑠 → 𝐿𝑎𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 ∶ η =
𝐹
𝑃
𝐴 = 𝜋𝑟 2 = 𝜋 62 𝑚2 → 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑟𝑒𝑎𝑐ℎ𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑛
𝑈𝐹 =
𝐹 384 𝜋 𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑒𝑚𝑖𝑡𝑡𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒
𝐸𝑎𝑣 = = = 10.67 𝑙𝑢𝑥 → 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝐴 𝜋 62 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸) 𝑢𝑛𝑑𝑒𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑀𝐹 =
𝒊)𝑬𝒄𝒆𝒏𝒕𝒆𝒓 &𝑬𝒆𝒅𝒈𝒆 & 𝑬𝒂𝒗 𝒘𝒊𝒕𝒉 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒐𝒓 ∶ 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸)𝑢𝑛𝑑𝑒𝑟 𝑒𝑣𝑒𝑟𝑦 𝑡ℎ𝑖𝑛𝑔 𝑖𝑠 𝑐𝑙𝑒𝑎𝑛
1
ω=4𝜋 → 𝐷𝑒𝑝𝑒𝑟𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶ 𝐷𝐹 =
𝑀𝐹
𝐼 𝐼
→ 𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ
𝐹 = 𝐼 ∗ ω = 300 ∗ 4𝜋 = 1200 𝜋 𝑙𝑢𝑚𝑒𝑛𝑠 𝑑 ℎ

𝐹 = 1200 𝜋 ∗ 0.6 = 720 𝜋 𝑙𝑢𝑚𝑒𝑛𝑠

𝐹 720 𝜋 (𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑜𝑟 𝑑𝑖𝑟𝑒𝑐𝑡 𝑡ℎ𝑒
𝐸𝑎𝑣 = = 2
= 20 𝑙𝑢𝑥 = 𝐸𝑐𝑒𝑛𝑡𝑒𝑟 = 𝐸𝑒𝑑𝑔𝑒 𝑙𝑖𝑔ℎ𝑡 𝒖𝒏𝒊𝒇𝒐𝒓𝒎𝒍𝒚 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝐴 𝜋6
8) Two lamps L1 and L2 are hung at a height of 9 m from the floor level. The distance between the lamps is 1 m. Lamp L1 is of
500 CP. If the illumination on the floor vertically below this lamp is 20 lux, find the candle power of the lamp L2.
ℎ = 9 𝑚 , distance between the lamps =1 m , 𝐼1 = 500 𝐶𝑃
𝐴𝑟𝑐
, 𝐸𝐴 = 20 𝑙𝑢𝑥 → 𝑃𝑙𝑎𝑛𝑒 𝑎𝑛𝑔𝑙𝑒 ∶ 𝜃 =
𝑟
(𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
→ 𝐼2 =? ? 𝐴
→ 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑔𝑙𝑒 ∶ ω = 2 = 2𝜋 1 − cos
𝜃
(𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠)
𝑟 2
𝐹
→ 𝐿𝑢𝑚𝑖𝑛𝑜𝑢𝑠 𝑖𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∶ 𝐼 = (𝐿𝑢𝑚𝑒𝑛Τ𝑠𝑡𝑒𝑟𝑎𝑑𝑖𝑎𝑛𝑠) 𝑜𝑟 (𝑐𝑑)
𝐼1 𝐼2 ω
𝐸𝐴 = 2 + 2 𝑐𝑜𝑠 3 θ → 𝐶𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝐶. 𝑃 =
𝐼𝑢𝑚𝑒𝑛𝑠(𝐹)
(𝑐𝑑)
ℎ ℎ ω
𝐹
3 → 𝐼𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 (luminance): 𝐸 = (𝐿𝑢𝑚𝑒𝑛Τ𝑚2 ) 𝑜𝑟 (𝑙𝑢𝑥)
500 𝐼2 9 𝐼
𝐴
20 = 2 + 2 → 𝐵𝑟𝑖𝑔ℎ𝑡𝑛𝑒𝑠𝑠: 𝐿 = 𝑐𝑑 Τ𝑚2
9 9 92 + 12 𝐴 cos 𝜃
⇒ 𝑓𝑜𝑟 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑑𝑖𝑓𝑓𝑢𝑠𝑒 𝑠ℎ𝑝ℎ𝑒𝑟𝑎 ∶ 𝐸 = 𝜋 𝐿
𝐹
→ 𝑀𝑒𝑎𝑛 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑐𝑎𝑛𝑑𝑙𝑒 𝑝𝑜𝑤𝑒𝑟 ∶ 𝑀𝑆𝐶𝑃 = (𝑐𝑑)
500 𝐼2 ∗ 0.982 500 + 0.982 ∗ 𝐼2 4𝜋
20 = + = → 𝐿𝑎𝑚𝑝 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑦 ∶ η =
𝐹
81 81 81 𝑃
→ 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑟𝑒𝑎𝑐ℎ𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑝𝑙𝑎𝑛
(20 ∗ 81) − 500 𝑈𝐹 =
𝐼2 = = 1140.5 𝐶𝑃 𝑡𝑜𝑡𝑎𝑙 𝑙𝑢𝑚𝑒𝑛𝑠 𝑒𝑚𝑖𝑡𝑡𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒
0.982 → 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 ∶
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸) 𝑢𝑛𝑑𝑒𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑀𝐹 =
𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝐸)𝑢𝑛𝑑𝑒𝑟 𝑒𝑣𝑒𝑟𝑦 𝑡ℎ𝑖𝑛𝑔 𝑖𝑠 𝑐𝑙𝑒𝑎𝑛
1
→ 𝐷𝑒𝑝𝑒𝑟𝑐𝑖𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 ∶ 𝐷𝐹 =
𝑀𝐹
𝐼 𝐼
→ 𝐸 = 2 cos θ = 2 𝑐𝑜𝑠 3 θ
𝑑 ℎ