[Physics Class Notes] on Derivation of Mirror Formula Pdf for Exam

Ibn al-Haitham was a physicist who had described the theory of vision. Scientists used to call him the father of optics.

Mirror Equation Derivation is very common among the students. Many questions are asked from this section of various boards as well as entrance tests. Mirror Formula Proof is very easy. It can be explained as the relation between the distance of an object, the distance of an image, and the focal length of the mirror.

Facts on Spherical Mirror

Some facts are there that you must know about the spherical mirror:

The object distance(u) is the length between the object and the pole of the mirror.

Image distance(v) is the length between the image and the pole of the mirror. 

Focal Length(f) is the distance between the principal focus and the pole of the mirror.

[frac{1}{f}] = [frac{1}{V}] + [frac{1}{U}]  

Where, 

f = focal length of the mirror

V = the distance of the image

U = the distance of the object

Proof of Mirror Formula

The formula which gives the relation between object distance (u), image distance (v) and focal length is defined as Mirror Formula.

[frac{1}{V}] + [frac{1}{U}] = [frac{1}{f}]  

In Triangle XYZ  and Triangle X’Y’Z



Triangle XYZ ~Triangle X’Y’Z (XX similarity)              => XY /X’Y’ =XZ/X’Z —-(I) dfpe

Similarly, In PQRS  ~ X’Y’F

  

SR /X’Y’ = RQ/X’F

[frac{XY}{X’Y}] = [frac{RQ}{X’F}]   

[frac{XY}{X’Y}] = [frac{RQ}{X’F}]        (XY=SR)  —-(II)

From(i) &(ii)

[frac{XZ}{X’Z}] = [frac{RQ}{X’F}]     

[frac{XZ}{X’Z}] = [frac{RQ}{X’F}]     

=> 
[frac{X’Z}{XZ}] = [frac{X’F}{RQ}]       

=>
[frac{(ZR-X’R)}{(XR-ZR)}] = [frac{(X’R-RF)}{RF}]



Now, RF = -f ;  ZR = 2RF = -2f ; XR = -u ; and X’R = -v

Put these value in above relation:

[frac{(-2f)-(-v)}{(-u)-(-2f)}] = [frac{(-v)-(-F)}{-f}]


=> uv = fv +uf

=> 
[frac{1}{f}] = [frac{1}{U}] + [frac{1}{v}]  

Derive Mirror Formula for Convex

From the below diagram, you can get to know the Derivation of Mirror Formula for Convex Mirror.


The above picture shows the following value:

 -u = PB

+V = PB’

+b = PF

+R = PC

In triangle ABC and A’B’C

[frac{AB}{A’B}] = [frac{CB}{CB’}] → 1

In triangle ABP and A/B/P

[frac{AB}{A’B}] = [frac{PB}{PB’}] → 2

From 1 and 2, we get

[frac{CB}{CB’}] = [frac{PB}{PB’}] …. 3

We also know CB = PB + PC and CB’ = PC – PB’

Putting the above values in eq (3)

[frac{PB+PC}{PC-PB’}] = [frac{PB}{PB’}] = [frac{-u+R}{R-v}] = [frac{-u}{v}]

⇒ -uv + vR = -uR + uv

⇒ uR + vR = 2uv

After dividing, uvR The final form of the equation is

[frac{1}{V}] + [frac{1}{U}] = [frac{1}{f}]  

Derive Mirror Formula for Concave Mirror

The Derivation of Mirror Formula for Concave Mirror is shown hereunder:


From the above image, we get

[frac{B’A’}{PM}] = [frac{B’F}{FP}] or [frac{B’A’}{BA}] = [frac{B’F}{FP}] (∵ PM = AB)….1

 [frac{B’A’}{BA}] = [frac{B’P}{BP}] ( ∵A P B angle = A’ P B’ angle)….. 2  

Equating 1 and 2, we will get,

[frac{B’F}{FP}] = [frac{B’P-FP}{FP}] = [frac{B’P}{BP}]

Also, B’P = -v, FP = -f, BP = -u

[frac{-v+f}{-f}] = [frac{-v}{-u}]

or, [frac{v-f}{f}] =[frac{v}{u}]

or, [frac{1}{u}] + [frac{1}{v}] = [frac{1}{f}]  

Derivation of Mirror Formula for Convex Lens


In the above image, you can notice two similar triangles i.e. △ABO and △A’B’O.

We can write 

[frac{A’B’}{AB}] = [frac{OB’}{OB}]   (1)

Also, △A’B’F and △OCF are similar.

We can write a relation as

[frac{A’B’}{OC}] = [frac{FB’}{OF}]

However, OC = OB

Then, [frac{A’B’}{AB}] = [frac{FB’}{OF}]   (2)

Equating both the equation 1 and 2, we get

[frac{OB’}{OB}] = [frac{FB’}{OF}] = [frac{OB’-OF}{OF}]  

If we conduct some sign convention, we will find

  • OB=-u,  

  • OB’=v 

  • and OF=f

[frac{v}{-u}] = [frac{v-f}{f}]   

vf = -uv + uf or uv = uf – vf 

Dividing uvf into both sides, we get

[frac{uv}{uvf}] = [frac{uf}{uvf}] – [frac{vf}{uvf}]

⇒ [frac{1}{f}] = [frac{1}{v}] – [frac{1}{u}] (This is the convex lens formula)

Sign Conventions for Mirror Equation

Mirror Equation follows certain sign conventions given below:

  • In the rectangular coordinate system, the principal axis of the mirror is taken along the x-axis, and its pole is taken as the origin.

  • All the distances parallel to the principal axis of the mirror are measured from the pole of the mirror.

  • The object is taken on the left side of the mirror. Hence, light is incident on the mirror from the left-hand side.

  • The distances measured in the direction of the incident light are taken as positive.

  • The distances measured in the direction opposite to the direction of incident light are taken as negative.

  • The heights measured upwards and perpendicular to the principal axis of the mirror are taken as positive.

  • The heights measured downwards and perpendicular to the principal axis of the mirror are taken as negative. 

Solved Examples

Derive the relation u1+v1=R2 for a concave mirror

Solution:

The relationship between object distance (u), the image distance (v) and the focal length (f) of the mirror is known as the mirror formula.

Suppose, an object AB is placed at a distance u from the pole of the concave mirror of a small aperture, just beyond the center of curvature. Hence, its real, inverted and diminished image AB is formed at a distance v in front of the mirror.

According to the Cartesian sign convention,

Object distance (PB)=−u

Image distance (PB)=−v

Focal length (PF)=−f

Radius of curvature (PC)=−R

It is clear from the geometry of the figure, right-angled △ ABP and △ABP are similar.

∴ABA′B′=PBP′B′=u−v

​∴ABA′B′=uv ……..(i)

Similarly, △ ABC and △ A′B′C’ are similar.

∴ABA′B′​=CBC′B′ …….(ii)

From figure,

CB′PC′PB′=−R−(−v)=−R+v

and CB=PB−PC=−u−(−R)=−u+R

From (ii),

 ABA′B′​=−u+R−R+V …… (iii)

Comparing (i) and (iii),

uv=−u+R−R+v

∴−uv+Rv=−Ru+vu 

or, R(u+v)=2uv

∴v1+u1=R2

(Dividing both sides by Ruv)

Hence, Proved.

Conclusion

This article will help you to know about the mirror formula and its relevant usages. With the use of optics, we can calculate and analyze the behaviour of light and image formation. Thus, this chapter is very helpful.

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