Below are some representative answers to the ‘practice questions’ that appear within any chapters where we need to learn to apply some maths. They’re designed to help you focus your learning, so make sure you have a go at answering them yourself before you take a peek at the answers!

## Chapter 2

2.1.1 If an object is placed 12 cm in front of a surface, what is the object’s vergence at the point where light from the object meets the surface?

• l = −0.12

• n = 1.00

• L = n/l

• L = 1.00/ −0.12

• L = −8.33D

2.1.2 If an object is placed 7 cm in front of a surface, what is the object’s vergence at the point where light from the object meets the surface?

• l = −0.07

• n = 1.00

• L = n/l

• L = 1.00 / −0.07

• L = −14.29D

2.2.1 If a light ray is incident on a glass block (refractive index 1.523) at an angle of 35°, what is the angle of refraction?

• i = 35°

• n = 1.00

• n′ = 1.523

• n (sin i) = n′ (sin i′)

• 1.00 (sin 35) = 1.523 (sin i′)

• 0.5735../ 1.523 = (sin i′)

• 0.3766.. = (sin i′)

• sin −1 (0.3766..) = i′

• i′ = 22.12°

2.2.2 If a light ray is incident on a piece of plastic (refractive index 1.498) at an angle of 25°, what is the angle of refraction?

• i = 25°

• n = 1.00

• n′ = 1.498

• n (sin i) = n′ (sin i′)

• 1.00 (sin 25) = 1.498 (sin i′)

• 0.4226.. / 1.498 = (sin i′)

• 0.2821.. = (sin i′)

• sin −1 (0.2821..) = i′

• i′ = 16.39°

2.2.3 If a light ray refracts out of a glass block (refractive index 1.523) at an angle of 20°, what was the angle of incidence at the back surface?

• i′ = 20°

• n = 1.523 (because the light is leaving the glass block)

• n′ = 1.00 (because the light is moving into air)

• n (sin i) = n′ (sin i′)

• 1.523 (sin i) = 1.00 (sin 20)

• 1.523 (sin i) = 0.3420..

• (sin i) = 0.3420.. / 1.523

• (sin i) = 0.2246..

• i = sin −1 (0.2246..)

• i = 12.98°

2.3.1 Calculate the lateral displacement of a light ray that enters a 3-cm-wide glass block (refractive index 1.523) at an angle of 31°.

• i 1 = 31°

• n = 1.00

• n′ = 1.523

• d = 0.03

• n (sin i) = n′ (sin i′)

• 1.00 (sin 31) = 1.523 (sin i′)

• 0.5150../ 1.523 = (sin i′)

• 0.3381.. = (sin i′)

• sin −1 (0.3381..) = i′

• i′ = 19.77..°

• s = (d (sin (i 1 – i 1 ′))) / (cos (i 1 ′))

• s = (0.03 (sin(31 − 19.77..))) / (cos (19.77..))

• s = 0.006211 m

• s = 0.62 cm

2.3.2 Calculate the lateral displacement of a light ray that enters a 11-cm-wide glass block (refractive index 1.523) at an angle of 29°.

• i 1 = 29°

• n = 1.00

• n′ = 1.523

• d = 0.11

• n (sin i) = n′ (sin i′)

• 1.00 (sin 29) = 1.523 (sin i′)

• 0.4848.. / 1.523 = (sin i′)

• 0.3183.. = (sin i′)

• sin −1 (0.3183..) = i′

• i′ = 18.56..°

• s = (d (sin (i 1 – i 1 ′))) / (cos (i 1 ′))

• s = (0.11 (sin(29 − 18.56..))) / (cos (18.56..))

• s = 0.02102 m

• s = 2.10 cm

2.4.1 Determine the power of a convex spherical glass surface (refractive index 1.523) with a radius of curvature of 18 cm.

• r = +0.18 m

• n = 1.00

• n′ = 1.523

• F = (n′ – n) / r

• F = (1.523 – 1.00) / 0.18

• F = +2.91D

2.4.2 Determine the power of a concave spherical glass surface (refractive index 1.523) with a radius of curvature of 10 cm.

• r = −0.1 m

• n = 1.00

• n′ = 1.523

• F = (n′ – n) / r

• F = (1.523 – 1.00) / −0.1

• F = −5.23D

2.4.3 Determine the power of a convex spherical glass surface (refractive index 1.523) with a radius of curvature of 50 cm.

• r = +0.5 m

• n = 1.00

• n′ = 1.523

• F = (n′ – n) / r

• F = (1.523 – 1.00) / 0.5

• F = +1.05D

2.5.1 An object is placed 10 cm in front of a convex spherical glass surface (refractive index 1.523) with a radius of curvature of 8 cm. Where does the image form?

• l = −0.1 m

• r = +0.08 m

• n = 1.00

• n′ = 1.523

• F = (n′ – n) / r

• F = (1.523 – 1.00) / 0.08

• F = +6.54..D

• L = n / l

• L = 1.00 / −0.1

• L = −10

• L′ = L + F

• L′ = −10 + 6.54..

• L′ = −3.46..

• l′ = n′ / L′

• l′ = 1.523 / −3.46..

• l′ = −0.4399 m

• l′ = −43.99 cm

2.5.2 An object is placed 30 cm in front of a concave spherical glass surface (refractive index 1.523) with a radius of curvature of 15 cm. Where does the image form?

• l = −0.3 m

• r = −0.15 m

• n = 1.00

• n′ = 1.523

• F = (n′ – n) / r

• F = (1.523 – 1.00) / −0.15

• F = −3.49..D

• L = n / l

• L = 1.00 / −0.3

• L = −3.33..

• L′ = L + F

• L′ = −3.49.. + −3.33..

• L′ = −6.82

• l′ = n′ / L′

• l′ = 1.523 / −6.82

• l′ = −0.2233 m

• l′ = −22.33 cm

## Chapter 3

3.1.1 A biconcave thin lens has a front surface power of −4.00D and a back surface power of −8.00D. What is the overall power of this lens?

• F 1 = −4.00D

• F 2 = −8.00D

• F = F 1 + F 2

• F = −4.00 + −8.00

• F = −12.00D

3.1.2 A plus meniscus thin lens has a front surface power of +6.50D and a back surface power of −2.00D. What is the overall power of this lens?

• F 1 = +6.50D

• F 2 = −2.00D

• F = F 1 + F 2

• F = 6.50 + −2.00

• F = +4.50D

3.2.1 A biconvex thin lens has a focal length of 30 cm. What is its power?

• n′ = 1.00

• f′ = +0.30

• F = n′ / f′

• F = 1.00 / +0.3

• F = +3.33D

3.2.2 A biconcave thin lens has a focal length of 11 cm. What is its power?

• n′ = 1.00

• f′ = −0.11

• F = n′ / f′

• F = 1.00 / −0.11

• F = −9.09D

3.2.3 A biconvex thin lens has a focal length of 25 cm. What is its power in water (refractive index 1.333)?

• n′ = 1.333

• f′ = +0.25

• F = n′ / f′

• F = 1.333 / +0.25

• F = +5.32D

3.2.4 A biconcave thin lens has a power of −4.00D. What is its focal length?

• n′ = 1.00

• F = −4.00D

• f′ = n′ / F

• f′ = 1.00 / −4.00

• f′ = −0.25 m or −25.00 cm

3.3.1 An object is placed 50 cm in front of a biconvex thin lens with a power of +10.00D. Where does the image form?

• l = −0.50 m

• F = +10.00D

• n = 1.00

• n′ = 1.00

• L = n / l

• L = 1.00 / −0.50

• L = −2.00

• L′ = L + F

• L′ = −2.00 + +10.00

• L′ = +8.00

• l′ = n′ / L′

• l′ = 1.00 / +8.00

• l′ = +0.125 m or +12.50 cm (right of lens)

3.3.2 An object is placed 35 cm in front of a biconcave thin lens with a power of −12.00D. What is the magnification of the image?

• l = −0.35 m

• F = −12.00D

• n = 1.00

• n′ = 1.00

• L = n / l

• L = 1.00 / −0.35

• L = −2.86..

• L′ = L + F

• L′ = −2.86.. + −12.00

• L′ = −14.86..

• m = L / L′

• m = −2.86.. / −14.86..

• m = +0.19X

3.3.3 An object is placed 15 cm in front of a biconcave thin lens with a focal length of 20 cm. Where does the image form?

• l = −0.15 m

• f′ = −0.20 m

• n = 1.00

• n′ = 1.00

• F = n′ / f′

• F = 1.00 / −0.20

• F = −5.00D

• L = n / l

• L = 1.00 / −0.15

• L = −6.67..

• L′ = L + F

• L′ = −6.67.. + −5.00

• L′ = −11.67..

• l′ = n′/ L′

• l′ = 1.00 / −11.67

• l′ = −0.0857 m or −8.57 cm (left of lens)

3.4.1 Two thin lenses of powers +1.00D and +5.50D are in contact with each other. If an object is placed 10 cm in front of the first lens, where does the image form?

• l = −0.10 m

• F 1 = +1.00D

• F 2 = +5.50D

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2

• F e = +1.00 + +5.50

• F e = +6.50D

• L = n / l

• L = 1.00 / −0.10

• L = −10.00

• L′ = L + F e

• L′ = −10 + 6.50

• L′ = −3.50

• l′ = n′ / L′

• l′ = 1.00 / −3.50

• l′ = −0.2857 m or −28.57 cm (left of lenses)

3.4.2 Two thin lenses of powers +6.50D and −4.25D are in contact with each other. If an object is placed 40 cm in front of the first lens, what is the linear magnification?

• l = −0.40 m

• F 1 = +6.50D

• F 2 = −4.25D

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2

• F e = +6.50 + −4.25

• F e = +2.25D

• L = n / l

• L = 1.00 / −0.40

• L = −2.50

• L′ = L + F e

• L′ = −2.50 + 2.25

• L′ = −0.25

• m = L / L′

• m = −2.50 / −0.25

• m = +10.00X

3.4.3 Three thin lenses of powers +1.00D, −3.50D and +6.75D are in contact with each other. If an object is placed 25 cm in front of the first lens, where does the image form?

• l = −0.25 m

• F 1 = +1.00D

• F 2 = −3.50D

• F 3 = +6.75D

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2 + F 3

• F e = +1.00 + −3.50 + +6.75

• F e = +4.25D

• L = n / l

• L = 1.00 / −0.25

• L = −4.00

• L′ = L + F e

• L′ = −4+ 4.25

• L′ = +0.25

• l′ = n′ / L′

• l′ = 1.00 / +0.25

• l′ = +4.00 m or +400.00 cm (right of lenses)

3.5.1 Two thin lenses of powers +4.50D and −1.00D are separated by a distance of 5 cm. What is the back vertex power of the system?

• F 1 = +4.50D

• F 2 = −1.00D

• d = 0.05 m

• F v ′ = (F 1 + F 2 − dF 1 F 2 ) / (1 − dF 1 )

• F v ′ = (+4.50 + −1.00 – (0.05)(+4.50)(−1.00)) / (1 − (0.05)(+4.50))

• F v ′ = +4.81D

3.5.2 Two thin lenses of powers +8.50D and −9.00D are separated by a distance of 11 cm. What is the front vertex power of the system?

• F 1 = +8.50D

• F 2 = −9.00D

• d = 0.11 m

• F v = (F 1 + F 2 − dF 1 F 2 ) / (1 – dF 2 )

• F v = (+8.50 + −9.00 – (0.11)(+8.50)(−9.00)) / (1 − (0.11)(−9.00))

• F v = +3.98D

3.6.1 Two thin lenses of powers +5.75D and −3.00D are separated by a distance of 8 cm. What is the back vertex focal length of the system?

• F 1 = +5.75D

• F 2 = −3.00D

• d = 0.08 m

• F v ′ = (F 1 + F 2 − dF 1 F 2 ) / (1 − dF 1 )

• F v ′ = (+5.75 + −3.00 – (0.08)(+5.75)(−3.00)) / (1 − (0.08)(+5.75))

• F v ′ = +7.65..D

• f v ′ = n / F v

• f v ′ = 1 / +7.65..

• f v ′ = +0.1308 m or +13.08 cm

3.6.2 Two thin lenses of powers −6.50D and −2.00D are separated by a distance of 11 cm. What is the front vertex focal length of the system?

• F 1 = −6.50D

• F 2 = −2.00D

• d = 0.11 m

• F v = (F 1 + F 2 − dF 1 F 2 ) / (1 – dF 2 )

• F v = (−6.50 + −2.00 – (0.11)(−6.50)(−2.00)) / (1 − (0.11)(−2.00))

• F v = −8.14..

• f v = −(n / F v )

• f v = −(1 / −8.14…)

• f v = − 0.1229 m or −12.29 cm

3.7.1 Two thin lenses of powers +5.00D and +1.00D are separated by a distance of 15 cm. If an object is placed 25 cm in front of the first lens, where will the image form?

• F 1 = +5.00D

• F 2 = +1.00D

• d = 0.15 m

• l 1 = −0.25 m

• n = 1.00

• n′ = 1.00

• L 1 = n / l 1

• L 1 = 1.00 / −0.25

• L 1 = −4.00

• L 1 ′ = L 1 + F 1

• L 1 ′ = −4.00 + +5.00

• L 1 ′ = +1.00

• L 2 = L 1 ′ / (1 – dL 1 ′)

• L 2 = +1.00 / (1 – (0.15 × 1.00))

• L 2 = +1.176…

• L 2 ′ = L 2 + F 2

• L 2 ′ = +1.176… + 1.00

• L 2 ′ = +2.176…

• l 2 ′ = n′ / L 2

• l 2 ′ = 1.00 / +2.176…

• l 2 ′ = +0.4595 m or +45.95 cm

3.7.2 Two thin lenses of powers −2.50D and +8.25D are separated by a distance of 8 cm. If an object is placed 10 cm in front of the first lens, where will the image form?

• F 1 = −2.50D

• F 2 = +8.25D

• d = 0.08 m

• l 1 = −0.1 m

• n = 1.00

• n′ = 1.00

• L 1 = n / l 1

• L 1 = 1.00 / −0.1

• L 1 = −10.00

• L 1 ′ = L 1 + F 1

• L 1 ′ = −10.00 + −2.50

• L 1 ′ = −12.50

• L 2 = L 1 ′ / (1 – dL 1 ′)

• L 2 = −12.50 / (1 – (0.08×−12.50))

• L 2 = −6.25

• L 2 ′ = L 2 + F 2

• L 2 ′ = −6.25 + 8.25

• L 2 ′ = +2.00

• l 2 ′ = n′ / L 2

• l 2 ′ = 1.00 / +2.00

• l 2 ′ = +0.50 m or +50.00 cm

3.8.1 Two thin lenses of powers +6.00D and +1.75D are separated by a distance of 10 cm. What is the equivalent power?

• F 1 = +6.00D

• F 2 = +1.75D

• d = 0.1 m

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (+6.00) + (+1.75) – (0.1 × (+6.00) × (+1.75))

• F e = +6.70D

3.8.2 Two thin lenses of powers −2.25D and +8.00D are separated by a distance of 11.5 cm. What is the equivalent power?

• F 1 = −2.25D

• F 2 = +8.00D

• d = 0.115 m

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (−2.25) + (+8.00) – (0.115 × (−2.25) × (+8.00))

• F e = +7.82D

3.9.1 Two thin lenses of powers −3.50D and +6.75D are separated by a distance of 10 cm. What is the secondary equivalent focal length of the system?

• F 1 = −3.50D

• F 2 = +6.75D

• d = 0.1 m

• n = 1.00

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (−3.50) + (+6.75) – (0.1 × (−3.50) × (+6.75))

• F e = +5.61..

• f e ′ = n / F e

• f e ′ = 1 / +5.61..

• f e ′ = +0.1782 m or +17.82 cm

3.9.2 Two thin lenses of powers −2.25D and +10.00D are separated by a distance of 50 cm. What is the primary equivalent focal length of the system?

• F 1 = −2.25D

• F 2 = +10.00D

• d = 0.5 m

• n = 1.00

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (−2.25) + (+10.00) – (0.5 × (−2.25) × (+10))

• F e = +19.00

• f e = −(n / F e )

• f e = −(1 / 19..)

• f e = −0.0526 m or −5.26 cm

3.9.3 A multiple lens system has a secondary equivalent focal length of +25 cm. What is the primary equivalent focal length of the system?

• f e ′ = +0.25 m

• f e = − f e

• f e = −0.25 m or −25.00 cm

3.10.1 Two thin lenses of powers −4.00D and +10.00D are separated by a distance of 5 cm. If an object is placed 25 cm in front of the primary focal point of the system, where will the image form?

• F 1 = −4.00D

• F 2 = +10.00D

• d = 0.05 m

• x = −0.25 m

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (−4.00) + (+10.00) – (0.05 × (−4.00) × (+10.00))

• F e = +8

• f e ′ = n / F e

• f e ′ = 1 / +8

• f e ′ = +0.125 m

• (f e ′) 2 = −xx′

• (0.125) 2 = − (−0.25) x′

• 0.0156.. / −0.25 = −x′

• −0.0625 = −x′

• x′ = +0.0625 m or +6.25 cm

3.10.2 Two thin lenses of powers +3.25D and +2.00D are separated by a distance of 14 cm. If an object is placed 16 cm in front of the primary focal point of the system, where will the image form?

• F 1 = +3.25D

• F 2 = +2.00D

• d = 0.14 m

• x = −0.16 m

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (+3.25) + (+2.00) – (0.14 × (+3.25) × (+2.00))

• F e = +4.34

• f e ′ = n / F e

• f e ′ = 1 / +4.34

• f e ′ = + 0.2304..m

• (f e ′) 2 = −xx′

• (0.2304..) 2 = − (−0.16) x′

• 0.0531.. / −0.16 = −x′

• −0.3318 = −x′

• x′ = +0.3318 m or +33.18 cm

3.10.3 A multiple lens system has a secondary equivalent focal length of +23.25 cm. If an object is placed 10 cm in front of the primary focal point of the system, where will the image form?

• f e = +0.2325 m

• x = −0.1 m

• n = 1.00

• n′ = 1.00

• (f e ′) 2 = −xx′

• (0.2325) 2 = − (−0.1) x′

• 0.0541.. / −0.1 = −x′

• −0.5406 = −x′

• x′ = +0.5406 m or +54.06 cm

3.11.1 Two thin lenses of powers −2.50D and +3.00D are separated by a distance of 3 cm. If an object is placed 45 cm in front of the primary focal point of the system, what is the linear magnification of the image?

• F 1 = −2.50D

• F 2 = +3.00D

• d = 0.03 m

• x = −0.45 m

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (−2.50) + (+3.00) – (0.030 × (−2.50) × (+3.00))

• F e = +0.725

• f e ′ = n / F e

• f e ′ = 1 / +0.725

• f e ′ = +1.38..m

• (f e ′) 2 = −xx′

• (1.38..) 2 = − (−0.45) x′

• 1.902.. / −0.45 = −x′

• −4.23.. = −x′

• x′ = +4.23..m

• m = −x′F e

• m = − (+4.23..)(+0.725)

• m = −3.07X

3.11.2 Two thin lenses of powers +3.25D and +7.00D are separated by a distance of 19 cm. If an object is placed 8 cm in front of the primary focal point of the system, what is the linear magnification of the image?

• F 1 = +3.25D

• F 2 = +7.00D

• d = 0.19 m

• x = −0.08 m

• n = 1.00

• n′ = 1.00

• F e = F 1 + F 2 – (dF 1 F 2 )

• F e = (+3.25) + (+7) – (0.19 × (+3.25) × (+7.00))

• F e = +5.93..

• f e ′ = n / F e

• f e ′ = 1 / +5.93..

• f e ′ = +0.1687..m

• (f e ′) 2 = −xx′

• (0.1687..) 2 = − (−0.08) x′

• 0.0284.. / −0.08 = −x′

• −0.3558.. = −x′

• x′ = +0.3558..m

• m = −x′F e

• m = − (+0.3558..)(+5.93..)

• m= −2.11x

## Chapter 4

4.1.1 A 7-cm-wide planoconvex lens has a radius of curvature of 12 cm. What is the sag of the lens?

• y = 0.035 m

• r = 0.12

• s = r − √(r 2 – y 2 )

• s = 0.12 − √(0.12 2 – 0.035 2 )

• s = 0.005218 m

• s = 0.52 cm

4.1.2 A 4-cm-wide planoconvex lens has a radius of curvature of 50 cm. What is the sag of the lens?

• y = 0.02 m

• r = 0.5

• s = r − √(r 2 – y 2 )

• s = 0.5 − √(0.5 2 – 0.02 2 )

• s = 0.0004 m

• s = 0.04 cm

4.1.3 A 5-cm-wide planoconvex lens has a radius of curvature of 10 cm. What is the sag of the lens?

• y = 0.025 m

• r = 0.1

• s = r − √(r 2 – y 2 )

• s = 0.1 − √(0.1 2 – 0.025 2 )

• s = 0.00318 m

• s = 0.32 cm

4.2.1 A thick lens (refractive index 1.523) has a convex front surface with a radius of curvature of 25 cm. What is the power of the front surface?

• n 1 ′ = 1.523

• r 1 = +0.25 m

• n 1 = 1.00

• F 1 = (n 1 ′ – n 1 ) / r 1

• F 1 = (1.523 – 1.00) / 0.25

• F 1 = +2.09D

4.2.2 A thick lens (refractive index 1.523) has a concave back surface with a radius of curvature of 25 cm. What is the power of the back surface?

• n 2 ′ = 1.000

• r 2 = +0.25 m

• n 2 = 1.523

• F 1 = (n 2 ′ – n 2 ) / r 2

• F 1 = (1.00 – 1.523) / +0.25

• F 1 = −2.09D

4.2.3 A 5-cm-thick lens (refractive index 1.523) has a front surface power of −2.00D and a back surface power of +4.00D. What is the power of the lens?

• t = 0.05 m

• n g = 1.523

• F 1 = −2.00D

• F 2 = +4.00D

• n = 1.00

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-1-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= t / n g

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-2-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.05 / 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-3-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.0328..

• F e = F 1 + F 2 <SPAN role=presentation tabIndex=0 id=MathJax-Element-4-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 1 F 2

• F e = −2.00 + 4.00 – (0.0328… × −2.00 × 4.00)

• F e = +2.26D

4.3.1 Imagine a 0.8 cm thick biconvex lens (surface powers +1.00D and +3.50D; refractive index 1.523). Where is the secondary principal plane, relative to the back surface of the lens?

• F 1 = +1.00D

• F 2 = +3.50D

• t = 0.008 m

• n = 1.00

• n g = 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-5-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= t / n g

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-6-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.008 / 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-7-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.0053..

• F v ′ = (F 1 + F 2 <SPAN role=presentation tabIndex=0 id=MathJax-Element-8-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 1 F 2 ) / (1− <SPAN role=presentation tabIndex=0 id=MathJax-Element-9-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 1 )

• F v ′ = (1.00 + 3.5 − 0.0053..×1.00×3.5) / (1 − (0.0053..× 1.00)

• F v ′ = +4.51..

• f v ′ = n / F v

• f v ′ = 1.00 / +4.51..

• f v ′ = +0.2220..

• F e = F 1 + F 2 <SPAN role=presentation tabIndex=0 id=MathJax-Element-10-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 1 F 2

• F e = 1.00+3.5 − 0.0053..×1.00×3.5

• F e = +4.48..

• f e ′ = n / F e

• f e ′ = 1.00 / +4.48..

• f e ′ = +0.2231..

• A 2 P′ = e′ = f v ′ – f e

• A 2 P′ = e′ = +0.2220.. − +0.2231..

• A 2 P′ = e′ = −0.0012 m

• Or 0.12 cm left of the back lens

4.3.2 Imagine a 1-cm-thick biconvex lens (surface powers +2.00D and +6.25D; refractive index 1.523). Where is the primary principal plane, relative to the back surface of the lens?

• F 1 = +2.00D

• F 2 = +6.25D

• t = 0.01 m

• n = 1.00

• n g = 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-11-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= t / n g

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-12-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.01 / 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-13-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.0066..

• F v = (F 1 +F 2 <SPAN role=presentation tabIndex=0 id=MathJax-Element-14-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 1 F 2 ) / (1 − <SPAN role=presentation tabIndex=0 id=MathJax-Element-15-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 2 )

• F v = (2.00 + 6.25 − 0.0066..×2×6.25) / (1 − (0.0066..×6.25)

• F v = +8.52..

• f v = −(n / F v )

• f v = −(1.00 / +8.52..)

• f v = −0.1174..

• F e = F 1 +F 2 <SPAN role=presentation tabIndex=0 id=MathJax-Element-16-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
F 1 F 2

• F e = 2.00 + 6.25 − 0.0053..×2.00×6.25

• F e = +8.17..

• f e = −(n / F e )

• f e = −(1.00 / +8.17..)

• f e = −0.1224..

• A 1 P = e = f v – f e

• A 1 P = e = −0.1174.. − −0.1224..

• A 1 P = e = +0.0050 m

• Primary principal plane is +0.5 cm right of front surface and is therefore −0.5 cm left of the back surface of the lens.

4.4.1 An object is placed 10 cm in front of a 4-cm-thick biconvex lens (refractive index 1.523) with a front surface power of +4.00D and a back surface power of +5.00D. Where does the image form relative to the back surface of the lens?

• l 1 = −0.10 m

• F 1 = +4.00D

• F 2 = +5.00D

• t = 0.04 m

• n = 1.00

• n′ = 1.523

• L 1 = n 1 / l 1

• L 1 = 1.00 / −0.10

• L 1 = −10.00

• L 1 ′ = L 1 + F 1

• L 1 ′ = −10.00 + 4.00

• L 1 ′ = −6.00

• l 1 ′ = n 1 ′ / L 1

• l 1 ′ = 1.523 / −6.00

• l 1 ′ = −0.2538..

• l 2 = l 1 ′ – t

• l 2 = −0.2538… – 0.04

• l 2 = −0.2938…

• L 2 = n 2 / l 2

• L 2 = 1.523 / −0.2938…

• L 2 = −5.18…

• L 2 ′ = L 2 + F 2

• L 2 ′ = −5.18… + 5.00

• L 2 ′ = −0.1832…

• l 2 ′ = n 2 ′ / L 2

• l 2 ′ = 1.00 / −0.1832…

• l 2 ′ = −5.4582 m or −545.82 cm

4.4.2 An object is placed 20 cm in front of a 1.5-cm-thick biconcave lens (refractive index 1.523) with a front surface power of −2.50D and a back surface power of −3.00D. Where does the image form relative to the back surface of the lens?

• l 1 = −0.20 m

• F 1 = −2.50D

• F 2 = −3.00D

• t = 0.015 m

• n = 1.00

• n′ = 1.523

• L 1 = n 1 / l 1

• L 1 = 1.00 / −0.20

• L 1 = −5.00

• L 1 ′ = L 1 + F 1

• L 1 ′ = −5.00 + −2.50

• L 1 ′ = −7.50

• l 1 ′ = n 1 ′ / L 1

• l 1 ′ = 1.523 / −7.50

• l 1 ′ = −0.2031..

• l 2 = l 1 ′ – t

• l 2 = −0.2031… – 0.015

• l 2 = −0.2181…

• L 2 = n 2 / l 2

• L 2 = 1.523 / −0.2181…

• L 2 = −6.98…

• L 2 ′ = L 2 + F 2

• L 2 ′ = −6.98… + −3.00

• L 2 ′ = −9.98…

• l 2 ′ = n 2 ′ / L 2

• l 2 ′ = 1.00 / −9.98…

• l 2 ′ = −0.1002 m or −10.02 cm

4.5.1 An object is placed 11 cm in front of a 4-cm-thick biconvex lens (refractive index 1.523) with a front surface power of +2.00D and a back surface power of +5.00D. Where does the image form relative to the back surface of the lens?

• l 1 = −0.11 m

• F 1 = +2.00D

• F 2 = +5.00D

• t = 0.04 m

• n = 1.00

• n′ = 1.523

• L 1 = n 1 / l 1

• L 1 = 1.00 / −0.11

• L 1 = −9.09..

• L 1 ′ = L 1 + F 1

• L 1 ′ = −9.09.. + 2

• L 1 ′ = −7.09..

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-17-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= t / n g

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-18-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.04 / 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-19-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.026..

• L 2 = L 1 ′ / (1 – <SPAN role=presentation tabIndex=0 id=MathJax-Element-20-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
L 1 ′)

• L 2 = −7.09../ (1 – (0.026.. × −7.09..))

• L 2 = −5.98..

• L 2 ′ = L 2 + F 2

• L 2 ′ = −5.98… + 5.00

• L 2 ′ = −0.98…

• l 2 ′ = n 2 ′ / L 2

• l 2 ′ = 1.00 / −0.98…

• l 2 ′ = −1.0229 m or −102.29 cm

4.5.2 An object is placed 25 cm in front of a 3-cm-thick biconcave lens (refractive index 1.523) with a front surface power of −1.50D and a back surface power of −5.00D. Where does the image form relative to the back surface of the lens?

• l 1 = −0.25 m

• F 1 = −1.50D

• F 2 = −5.00D

• t = 0.03 m

• n = 1.00

• n′ = 1.523

• L 1 = n 1 / l 1

• L 1 = 1.00 / −0.25

• L 1 = −4.00

• L 1 ′ = L 1 + F 1

• L 1 ′ = −4.00 + −1.50

• L 1 ′ = −5.50

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-21-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= t / n g

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-22-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.03 / 1.523

• <SPAN role=presentation tabIndex=0 id=MathJax-Element-23-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
= 0.020..

• L 2 = L 1 ′ / (1 – <SPAN role=presentation tabIndex=0 id=MathJax-Element-24-Frame class=MathJax style="POSITION: relative" data-mathml='t¯’>t
t ¯
L 1 ′)

• L 2 = −5.50/ (1 – (0.020..×−5.50))

• L 2 = −4.96..

• L 2 ′ = L 2 + F 2

• L 2 ′ = −4.96… + −5.00

• L 2 ′ = −9.96…

• l 2 ′ = n 2 ′ / L 2

• l 2 ′ = 1.00 / −9.96…

• l 2 ′ = −0.1004 m or −10.04 cm

## Chapter 6

6.1.1 Two plane mirrors are inclined at an angle of 35° towards one another. What will the angle of deviation be?

• a = 35°

• d = 360 − 2a

• d = 360 – 2(35)

• d = 360 – 70

• d = 290°

6.1.2 Two plane mirrors are inclined at an angle of 41.5° towards one another. What will the angle of deviation be?

• a = 41.5°

• d = 360 − 2a

• d = 360 – 2(41.5)

• d = 360 – 83

• d = 277°

6.2.1 A convex spherical mirror has a radius of curvature of 100 cm. What is its power?

• r = +1 m

• n = 1.00

• f = r / 2

• f = 1/ 2

• f = +0.5 m

• F = − (n / f)

• F = − (1.00 / +0.5)

• F = −2.00D

6.2.2 A concave spherical mirror has a focal length of 50 cm. What is its power?

• f = −0.5 m

• n = 1.00

• F = − (n / f)

• F = − (1.00 / −0.5)

• F = +2.00D

6.3.1 An object is placed 10 cm in front of a convex spherical mirror with a radius of curvature of 20 cm. Where does the image form?

• l = −0.1 m

• r = +0.20 m

• n = 1.00

• f = r / 2

• f = +0.20 / 2

• f = +0.10

• F = − (n / f)

• F = − (1.00 /+0.10)

• F = −10.00D

• L = n / l

• L = 1.00 / −0.1

• L = −10

• L′ = L + F

• L′ = −10 + −10

• L′ = −20

• l′ = − (n′ / L′)

• l′ = − (1.00 / −20)

• l′ = +0.05 m or +5.00 cm

6.3.2 An object is placed 30 cm in front of a concave spherical mirror with a power of +5.00D. Where does the image form?

• l = −0.3 m

• F = +5.00D

• n = 1.00

• L = n / l

• L = 1.00 / −0.3

• L = −3.33..

• L′ = L + F

• L′ = −3.33.. + −5

• L′ = +1.67..

• l′ = − (n′ / L′)

• l′ = − (1.00 / 1.67..)

• l′ = −0.60 m or −60.00 cm

6.4.1 An object is placed 20 cm in front of a convex spherical mirror with a radius of curvature of 30 cm. Where does the image form?

• l = −0.20 m

• r = +0.30 m

• n = 1.00

• (1 / l′) + (1 / l) = (2 / r)

• (1 / l′) + (1 / −0.20) = (2 / 0.3)

• (1 / l′) = 11.67..

• l′ = 1 / 11.67..

• l′ = +0.0857 m or +8.57 cm

6.4.2 An object is placed 25 cm in front of a concave spherical mirror with a power of +5.00D. Where does the image form?

• l = −0.25 m

• F = +5.00D

• n = 1.00

• f = − (n / F)

• f = − (1.00 / +5.00)

• f = −0.20

• (1 / l′) + (1 / l) = (1 / f)

• (1 / l′) + (1 / −0.25) = (1 / −0.20)

• (1 / l′) = −1

• l′ = 1 / −1

• l′ = −1.00 m or −100.00 cm

6.5.1 An object is placed 20 cm in front of a concave spherical mirror with a radius of curvature of 60 cm. What is the magnification of the image?

• l = −0.20 m

• r = −0.60 m

• n = 1.00

• (1 / l′) + (1 / l) = (2 / r)

• (1 / l′) + (1 / −0.20) = (2 / −0.60)

• (1 / l′) = 1.67..

• l′ = 1 / 1.67..

• l′ = +0.60 m

• m = − (l′ / l)

• m = − (+0.6 / −0.2)

• m = +3.00X (magnified, upright, virtual)

6.5.2 An object is placed 30 cm in front of a concave spherical mirror of power +10.00D. What is the magnification of the image?

• l = −0.30 m

• F = +10.00D

• n = 1.00

• f = − (n / F)

• f = − (1.00 / +10.00)

• f = −0.10

• (1 / l′) + (1 / l) = (1 / f)

• (1 / l′) + (1 / −0.30) = (1 / −0.10)

• (1 / l′) = −6.67..

• l′ = 1 / −6.67..

• l′ = −0.15 m

• m = − (l′ / l)

• m = − (−0.15 / −0.3)

• m = −0.50X (minified, inverted, real)

## Chapter 8

8.1.1 What is the chromatic aberration of a crown glass lens with a power of +20.00D?

• V = 60

• F = +20.00D

• CA = F / V

• CA = 20.00 / 60

• CA = 0.33

8.1.2 What is the chromatic aberration of a flint glass lens with a power of +10.00D?

• V = 30

• F = +10.00D

• CA = F / V

• CA = 10.00 / 30

• CA = 0.33

8.2.1 What power of flint glass would be required to remove chromatic aberration from a lens made partly of crown glass (+8.00D)?

• V c = 60

• V f = 30

• F c = +8.00D

• CA = 0

• CA = (F c / V c ) + (F f / V f )

• 0 = (8.00 / 60) + (F f / 30)

• −0.133.. = F f / 30

• −0.133.. × 30 = F f

• −4.00D = F f

8.2.2 What power of crown glass would be required to remove chromatic aberration from a lens made partly of flint glass (+2.00D)?

• V c = 60

• V f = 30

• F f = +2.00D

• CA = 0

• CA = (F c / V c ) + (F f / V f )

• 0 = (F c / 60) + (2.00 / 30)

• −0.067.. = F c / 60

• −0.067.. × 60 = F c

• −4.00D = F f

## Chapter 9

9.1.1 A ray of light is incident upon a prism at an angle of 20° and leaves the prism at an angle of 40°. If the prism has an apical angle of 55°, what is the angle of deviation of the light?

• i 1 = 20°

• i 2 ′ = 40°

• a = 55°

• d = (i 1 + i 2 ′) − a

• d = (20 + 40) − 55

• d = 5°

9.1.2 A ray of light is incident upon a prism at an angle of 21.2° and leaves the prism at an angle of 58.5°. If the prism has an apical angle of 30°, what is the angle of deviation of the light?

• i 1 = 21.2°

• i 2 ′ = 58.5°

• a = 30°

• d = (i 1 + i 2 ′) − a

• d = (21.2 + 58.5) − 30

• d = 49.7°

9.2.1 Calculate the minimum angle of deviation of a prism constructed of refractive index 1.498 of apical angle 60°.

• n p = 1.498

• n s = 1.00

• a = 60

• n s (sin0.5(a + d min )) = n p (sin (0.5×a))

• 1.00 (sin0.5(60 + d min )) = 1.498 (sin (0.5×60))

• sin0.5(60 + d min ) = 1.498 (sin (0.5×60))

• sin0.5(60 + d min ) = 0.749

• 0.5(60 + d min ) = sin −1 (0.749)

• 0.5(60 + d min ) = 48.50..

• 30 + 0.5(d min ) = 48.50..

• 0.5(d min ) = 48.50.. –30

• 0.5(d min ) = 18.50..

• d min = 18.50.. / 0.5

• d min = 37.01°

9.2.2 Calculate the minimum angle of deviation of a prism constructed of refractive index 1.65 of apical angle 30°.

• n p = 1.65

• n s = 1.00

• a = 30

• n s (sin0.5(a + d min )) = n p (sin (0.5×a))

• 1.00 (sin0.5(30 + d min )) = 1.65 (sin (0.5×30))

• sin0.5(30 + d min ) = 1.65 (sin (0.5×30))

• sin0.5(30 + d min ) = 0.427..

• 0.5(30 + d min ) = sin −1 (0.427..)

• 0.5(30 + d min ) = 25.28..

• 15 + 0.5(d min ) = 25.28..

• 0.5(d min ) = 25.28.. –15

• 0.5(d min ) = 10.28..

• d min = 10.28.. / 0.5

• d min = 20.56°

9.3.1 Determine the smallest angle of incidence at the first face of an equilateral triangular prism of refractive index 1.498 for light to just pass through the second face. The apical angle of the prism is 50°.

• n p = 1.498

• n s = 1.00

• a = 50

• sin(i c ) = 1 / n p

• sin(i c ) = 1 / 1.498

• i c = sin −1 (1 / 1.498)

• i c = 41.88..°

• i 2 = i c

• i 1 ′ = a − i 2

• i 1 ′ = 50 – 41.88..

• i 1 ′ = 8.12..°

• n s (sin i 1 ) = n p (sin i 1 ′)

• 1.00 (sin i 1 ) = 1.498 (sin 8.12..)

• sin i 1 = 1.498 (sin 8.12..) / 1

• i 1 = sin −1 (1.498 (sin 8.12..))

• i 1 = 12.22°

9.3.2 Determine the smallest angle of incidence at the first face of an equilateral triangular prism of refractive index 1.523 for light to just pass through the second face. The apical angle of the prism is 55°.

• n p = 1.523

• n s = 1.00

• a = 55

• sin(i c ) = 1 / n p

• sin(i c ) = 1 / 1.523

• i c = sin −1 (1 / 1.523)

• i c = 41.04..°

• i 2 = i c

• i 1 ′ = a − i 2

• i 1 ′ = 55 – 41.048..

• i 1 ′ = 13.96..°

• n s (sin i 1 ) = n p (sin i 1 ′)

• 1.00 (sin i 1 ) = 1.523 (sin 13.96..)

• sin i 1 = 1.523 (sin 13.96..) / 1

• i 1 = sin −1 (1.523 (sin 13.96..))

• i 1 = 21.55°

9.3.3 Determine the smallest angle of incidence at the first face of an equilateral triangular prism of refractive index 1.65 for light to just pass through the second face. The apical angle of the prism is 45°.

• n p = 1.65

• n s = 1.00

• a = 45

• sin(i c ) = 1 / n p

• sin(i c ) = 1 / 1.65

• i c = sin −1 (1 / 1.65)

• i c = 37.31..°

• i 2 = i c

• i 1 ′ = a − i 2

• i 1 ′ = 45 – 37.31..

• i 1 ′ = 7.69..°

• n s (sin i 1 ) = n p (sin i 1 ′)

• 1.00 (sin i 1 ) = 1.65 (sin 7.69..)

• sin i 1 = 1.65 (sin 7.69..) / 1

• i 1 = sin −1 (1.65 (sin 7.69..))

• i 1 = 12.76°

9.4.1 A prism produces 5 m of displacement at 100 m. What is its power?

• x = 500 cm

• y = 10,000 cm

• P = (100)(x/y)

• P = (100)(500 / 10,000)

• P = 5

9.4.2 A prism produces 15 cm of displacement at 65 cm. What is its power?

• x = 15 cm

• y = 65 cm

• P = (100)(x/y)

• P = (100)(15 / 65)

• P = 23.08

9.5.1 How much prism power will be induced if a patient looks through their −2.50DS lens 2 mm left of the optical centre?

• F = −2.50DS

• c = 0.2 cm

• P = cF

• P = 0.2 × −2.50

• P = −0.5

9.5.2 How much prism power will be induced if a patient looks through their +4.25DS lens 1 mm right of the optical centre?

• F = +4.25DS

• c = 0.1 cm

• P = cF

• P = 0.1 × 4.25

• P = 0.43

## Chapter 10

10.1.1 A single slit is placed 40 cm in front of a wall. If the first and second maxima are 8 cm apart from one another, what is the angle of diffraction?

• L = 0.4 m

• y = 0.08 m

• sin ϕ = y / √(L 2 + y 2 )

• sin ϕ = 0.08 / √(0.4 2 + 0.08 2 )

• ϕ = sin −1 (0.1961..)

• ϕ = 11.31°

10.1.2 A single slit is placed 55 cm in front of a wall. If the first and second maxima are 18 cm apart from one another, what is the angle of diffraction?

• L = 0.55 m

• y = 0.18 m

• sin ϕ = y / √(L 2 + y 2 )

• sin ϕ = 0.18 / √(0.55 2 + 0.18 2 )

• ϕ = sin −1 (0.3110..)

• ϕ = 18.12°

## Chapter 12

12.1.1 Calculate the planar angle shown as ‘?’ in the following image:

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