1. Example: An object 3 cm high is placed 20 cm from (a) a convex and (b) a concave spherical mirror, each of 10 cm focal length. Determine the position and nature of the image in each case.
a) Convex
S = 20 cm
ho = 3 cm
f = -10 cm
1 𝑠 + 1 𝑠 ′ = 1 𝑓
→ 1 𝑠 ′ = 1 𝑓 − 1 𝑠
=− 1 10 𝑐𝑚 − 1 20 𝑐𝑚
→ 1 𝑠 ′ = 3 20 𝑐𝑚
→ 𝑠 ′ =−6.67 cm
→ 𝑠 ′ =6.67 cm behind mirror (virtual image)
𝑚=− 𝑠 ′ 𝑠 =− −6.67 𝑐𝑚 20 𝑐𝑚
→𝑚=0.333 (image is upright)
ℎ 𝑖 =𝑚 ℎ 𝑜 = 𝑐𝑚 =1 𝑐𝑚 (image is upright)
b) Concave
S = 20 cm
ho = 3 cm
f = 10 cm
1 𝑠 + 1 𝑠 ′ = 1 𝑓
→ 1 𝑠 ′ = 1 𝑓 − 1 𝑠
= 1 10 𝑐𝑚 − 1 20 𝑐𝑚
→ 1 𝑠 ′ = 1 20 𝑐𝑚
→ 𝑠 ′ =20cm
→ 𝑠 ′ =20 cm in front of mirror (real image)
𝑚=− 𝑠 ′ 𝑠 =− 20 𝑐𝑚 20 𝑐𝑚
→𝑚=−1 (image is inverted)
ℎ 𝑖 =𝑚 ℎ 𝑜 = −1 3 𝑐𝑚 =−3 𝑐𝑚 (image is inverted)

2. ∆𝐶𝑃𝑂
𝜑+ 𝜃 1 =𝛼
∆𝐶𝑃𝐼
𝜑+ 𝜃 2 = 𝛼 ′
Using Snell’s Law and small angle approximations:
𝑛 1 𝑠𝑖𝑛 𝛼−𝜑 = 𝑛 2 𝑠𝑖𝑛 𝛼 ′ −𝜑
𝑛 1 𝛼−𝜑 = 𝑛 2 𝛼 ′ −𝜑
𝛼≈𝑠𝑖𝑛𝛼≈ ℎ 𝑠
𝛼 ′ ≈𝑠𝑖𝑛 𝛼 ′ ≈ ℎ 𝑠 ′
𝜑≈𝑡𝑎𝑛𝜑≈ ℎ 𝑅
𝑛 1 ℎ 𝑠 − ℎ 𝑅 = 𝑛 2 ℎ 𝑠 ′ − ℎ 𝑅
𝑛 1 𝑠 − 𝑛 2 𝑠 ′ = 𝑛 1 − 𝑛 2 𝑅
Using the sign convention for mirrors (Real is positive and virtual is negative – Note that a real image for a lens is to the right of the lens!) this can be generalized.
𝑛 1 𝑠 + 𝑛 2 𝑠 ′ = 𝑛 1 − 𝑛 2 𝑅

3. We can also look at the magnification:
𝑚= ℎ 𝑖 ℎ 𝑜 = 𝑛 1 𝑠 ′ 𝑛 2 𝑠
𝑛 1 𝑠𝑖𝑛 𝜃 1 = 𝑛 2 𝑠𝑖𝑛 𝜃 2
Using small angle approximation:
This can be generalized using the sign convention for mirrors:
𝑛 1 𝜃 1 = 𝑛 2 𝜃 2
𝜃 1 ≈𝑡𝑎𝑛 𝜃 1 ≈ ℎ 𝑜 𝑠
𝑚= ℎ 𝑖 ℎ 𝑜 =− 𝑛 1 𝑠 ′ 𝑛 2 𝑠
𝜃 2 ≈𝑡𝑎𝑛 𝜃 2 ≈ ℎ 𝑖 𝑠 ′

4. Example: A real object is positioned in air, 30 cm from a convex spherical surface of radius 5 cm. To the right of the interface, the refractive index is that of water (1.33).
a) Determine the location, size, orientation and type of image. Assume the medium extends very far to the right of the interface.
b) Repeat for the situation where the medium is 10 cm thick (at the center) and is bounded on the far side by a concave interface of radius 5 cm.