Introduction to Geometrical Optics: Math resources€¦ · Introduction to Geometrical Optics: Math...

Introduction to Geometrical Optics:

Math resources

Apratim MajumderOptics for Energy 2020

2020/09/03Thursday

Part II

Example Problem:Thick Lens

(Slides 38-40 in Lecture notes PDF)

𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 =

Refractionat

spherical surface (R2)

Free spacepropagation inside lens

(𝐷𝐷𝐷)

Refractionat


𝑛𝑛𝑖𝑖𝑖𝑖𝛼𝛼𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖𝑖𝑖

From Output side to Input side

Solution:Only consider ray entering and exiting lens(ignore Free Space Prop. In air outside lens)


Refractionat



(𝐷𝐷𝐷)

Refractionat



Here, 𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑛𝑛𝑖𝑖𝑖𝑖 = 𝐷




Refractionat



(𝐷𝐷𝐷)

Refractionat



Here, 𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑛𝑛𝑖𝑖𝑖𝑖 = 𝐷Hence:

𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 =

Refractionat



(𝐷𝐷𝐷)

Refractionat


𝛼𝛼𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖𝑖𝑖




Refractionat



(𝐷𝐷𝐷)

Refractionat





𝐷 0𝐷𝐷𝑛𝑛

𝐷

Free space propagation

matrix

𝐷 −𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜 − 𝑛𝑛𝑖𝑖𝑖𝑖

𝑅𝑅0 𝐷

Refraction at spherical surface

matrix𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 =

𝐷 −𝐷 − 𝑛𝑛′

𝑅𝑅20 𝐷

𝐷 0𝐷𝐷1𝑛𝑛′

𝐷𝐷 −

𝑛𝑛′ − 𝐷𝑅𝑅1

0 𝐷


Solve the equation:



𝑅𝑅20 𝐷


𝐷𝐷 −


0 𝐷


Solve the equation:



𝑅𝑅20 𝐷


𝐷𝐷 −


0 𝐷



𝐷𝑛𝑛′ − 𝐷𝑅𝑅2

0 𝐷

𝐷𝐷 − 𝑛𝑛′

𝑅𝑅1𝐷𝐷1𝑛𝑛′

𝐷𝐷1𝑛𝑛′

𝐷 − 𝑛𝑛′

𝑅𝑅1+ 𝐷


Solve the equation:



𝑅𝑅20 𝐷


𝐷𝐷 −


0 𝐷




0 𝐷





𝑅𝑅1+ 𝐷



𝐷 +𝐷𝐷1𝑛𝑛′



𝑅𝑅1+𝑛𝑛′ − 𝐷𝑅𝑅2

(𝐷𝐷1𝑛𝑛′


𝑅𝑅1+ 𝐷)




𝑅𝑅1+ 𝐷


Solve the equation:



𝑅𝑅20 𝐷


𝐷𝐷 −


0 𝐷




0 𝐷





𝑅𝑅1+ 𝐷









𝑅𝑅1+ 𝐷)




𝑅𝑅1+ 𝐷






𝑅𝑅1+𝐷𝐷1(𝐷 − 𝑛𝑛′)(𝑛𝑛′−𝐷)

𝑛𝑛′𝑅𝑅1𝑅𝑅2+𝑛𝑛′ − 𝐷𝑅𝑅2




𝑅𝑅1+ 𝐷


Solve the equation:



𝑅𝑅20 𝐷


𝐷𝐷 −


0 𝐷




0 𝐷





𝑅𝑅1+ 𝐷









𝑅𝑅1+ 𝐷)




𝑅𝑅1+ 𝐷






𝑅𝑅1+𝐷𝐷1(𝐷 − 𝑛𝑛′)(𝑛𝑛′−𝐷)

𝑛𝑛′𝑅𝑅1𝑅𝑅2+𝑛𝑛′ − 𝐷𝑅𝑅2




𝑅𝑅1+ 𝐷





−(𝑛𝑛′ − 𝐷){𝐷𝑅𝑅1

−𝐷𝑅𝑅2

+𝐷𝐷1(𝑛𝑛′−𝐷)𝑛𝑛′𝑅𝑅1𝑅𝑅2

}




𝑅𝑅1+ 𝐷





−(𝑛𝑛′ − 𝐷){𝐷𝑅𝑅1

−𝐷𝑅𝑅2

+𝐷𝐷1(𝑛𝑛′−𝐷)𝑛𝑛′𝑅𝑅1𝑅𝑅2

}




𝑅𝑅1+ 𝐷


In System Matrix, term −𝑀𝑀12 is the power term

Hence, Power of the lens = (𝑛𝑛′ − 𝐷){ 1𝑅𝑅1

− 1𝑅𝑅2

+ 𝐷𝐷1(𝑖𝑖′−1)

𝑖𝑖′𝑅𝑅1𝑅𝑅2

Also, Power of lens (𝑃𝑃) = 1/Focal length of lens (𝑓𝑓)and in the case of a thick lens, focal length is the Effective Focal Length (EFL)

Thus, EFL = 𝑓𝑓 and 1𝑓𝑓

= (𝑛𝑛′ − 𝐷){ 1𝑅𝑅1

− 1𝑅𝑅2

+ 𝐷𝐷1(𝑖𝑖′−1)

𝑖𝑖′𝑅𝑅1𝑅𝑅2

Example Problem:Imaging Condition(Slides 41-44 in Lecture notes PDF)

S S’

n n’


Free spacepropagation

(𝑆𝑆𝑆)

Refraction insidethe optical system


(𝑆𝑆)



Solution:The optical system may be complicated and have many lenses inside.Hence, simple consider a “black box” with 1st and 2nd Principal planes,

where the rays suffer refraction for the first and last times. Also, let this system have a power P that defines how the rays behave.

Complicated Optical System with power P

S S’

n n’



(𝑆𝑆𝑆)



(𝑆𝑆)


Here, 𝑛𝑛𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑛𝑛’ and 𝑛𝑛𝑖𝑖𝑖𝑖 = 𝑛𝑛Hence:

𝑛𝑛’𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜

=Free space

propagation(𝑆𝑆𝑆)

Refraction insidethe optical system with power P


(𝑆𝑆)

𝑛𝑛𝛼𝛼𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖𝑖𝑖


Solution:The optical system may be complicated and have many lenses inside.Hence, simple consider a “black box” with 1st and 2nd Principal planes,

where the rays suffer refraction for the first and last times. Also, let this system have a power P that defines how the rays behave.


S S’

n n’


=Free space

propagation(𝑆𝑆𝑆)



(𝑆𝑆)


From Output side to Input sideSolution:


S S’

n n’


=𝐷 0𝑆𝑆𝑆𝑛𝑛′

𝐷𝐷 −𝑃𝑃0 𝐷

𝐷 0𝑆𝑆𝑛𝑛

𝐷𝑛𝑛𝛼𝛼𝑖𝑖𝑖𝑖𝑥𝑥𝑖𝑖𝑖𝑖

Solve the equation:






Solve the equation:








𝐷

𝐷 −𝑃𝑃𝑆𝑆𝑛𝑛

−𝑃𝑃

𝑆𝑆𝑛𝑛

𝐷


Solve the equation:








𝐷


−𝑃𝑃

𝑆𝑆𝑛𝑛

𝐷



=𝐷 −

𝑃𝑃𝑆𝑆𝑛𝑛

−𝑃𝑃

𝑆𝑆′

𝑛𝑛′𝐷 −


+𝑆𝑆𝑛𝑛

−𝑃𝑃𝑆𝑆′

𝑛𝑛′+ 𝐷


Solve the equation:








𝐷


−𝑃𝑃

𝑆𝑆𝑛𝑛

𝐷



=𝐷 −


−𝑃𝑃

𝑆𝑆′

𝑛𝑛′𝐷 −


+𝑆𝑆𝑛𝑛

−𝑃𝑃𝑆𝑆′

𝑛𝑛′+ 𝐷



=𝐷 −


−𝑃𝑃

𝑆𝑆′

𝑛𝑛′+𝑆𝑆𝑛𝑛−𝑃𝑃𝑆𝑆𝑆𝑆′

𝑛𝑛𝑛𝑛′𝐷 −

𝑃𝑃𝑆𝑆′

𝑛𝑛′


System Matrix

Solution:


S S’

n n’


=𝐷 −


−𝑃𝑃

𝑆𝑆′


𝑛𝑛𝑛𝑛′𝐷 −

𝑃𝑃𝑆𝑆′

𝑛𝑛′


𝑛𝑛’𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜 = 𝐷 −𝑃𝑃𝑆𝑆𝑛𝑛

𝑛𝑛𝛼𝛼𝑖𝑖𝑖𝑖 − 𝑃𝑃𝑥𝑥𝑖𝑖𝑖𝑖

𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 =𝑆𝑆′


𝑛𝑛𝑛𝑛′𝑛𝑛𝛼𝛼𝑖𝑖𝑖𝑖 + (𝐷 −

𝑃𝑃𝑆𝑆′

𝑛𝑛′)𝑥𝑥𝑖𝑖𝑖𝑖

Equating the terms:


S S’

n n’

𝑛𝑛’𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜 = 𝐷 −𝑃𝑃𝑆𝑆𝑛𝑛

𝑛𝑛𝛼𝛼𝑖𝑖𝑖𝑖 − 𝑃𝑃𝑥𝑥𝑖𝑖𝑖𝑖

𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 =𝑆𝑆′


𝑛𝑛𝑛𝑛′𝑛𝑛𝛼𝛼𝑖𝑖𝑖𝑖 + (𝐷 −

𝑃𝑃𝑆𝑆′

𝑛𝑛′)𝑥𝑥𝑖𝑖𝑖𝑖

For proper imaging to take place, all rays from object point must meet at image point. Hence, 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 cannot have any angle (𝛼𝛼𝑖𝑖𝑖𝑖) dependence. Hence:

𝑆𝑆′

𝑖𝑖′+ 𝑆𝑆

𝑖𝑖− 𝑃𝑃𝑆𝑆𝑆𝑆

′

𝑖𝑖𝑖𝑖′= 0

or, 𝑛𝑛𝑆𝑆

+𝑛𝑛′

𝑆𝑆′= 𝑃𝑃 =

𝐷𝑓𝑓

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