Lens Formula


 
 
Concept Explanation
 

Lens Formula

If we place an object at a distance u from the optical centre 'O'  and its image is formed at a distance then

 frac{1}{v}-frac{1}{u}=frac{1}{f} 

This relationship is known as lens formula. Here is the focal length of this lens.

Proof:

Consider an object AB placed beyond 2F in front of a convex lens. On the other side of lens an image is formed between F and 2F which is real, inverted and smaller in size. Applying the sign conventions :

Object distance (OB ):= -u

Image distance(OB') := v

Focal length(OF): = +f

In Delta AOB andDelta A^IB^IO 

angle 1 = angle 2                      ( vertically opposite angle)

angle 3 =angle 4                       ( each 90^{circ})

therefore angle BAOsim angle B^IA^IO             (By AA similarity)

Therefore, Delta ABO and Delta A^IB^IO  are equi-angular and hence they are similar.

frac{A^IB^I}{AB}= frac{OB^I}{OB}=frac{+v}{-u}                                   ............[ 1 ]

Now  Delta OCF andDelta A^IB^IF 

angle 6 = angle 5                      ( vertically opposite angle)

angle COF =angle 4                       ( each 90^{circ})

therefore angle OCFsim angle B^IA^IF             (By AA similarity)

Therefore, Delta OCF and Delta A^IB^IF  are equi-angular and hence they are similar.

In particular , 

frac{A^IB^I}{OC}= frac{FB^I}{OF}

 But OC = AB

frac{A^IB^I}{AB}=frac{FB^I}{OF}

Since all distances are measured from optical centre , therefore,

frac{A^IB^I}{AB}=frac{FB^I}{OF}=frac{OB^I-OF}{OF}=frac{+v-(+f)}{+f}=frac{v-f}{f}                        ...........[ 2 ]

Now comparing eq. (1) and (2),

frac{+v}{-u}=frac{v-f}{f} 

Rightarrow fv=-u(v-f)

Rightarrow fv=-uv+fu

Dividing both the sides by uvf,

Rightarrowfrac{fv}{uvf}=-frac{uv+fu}{uvf}

 frac{}{}frac{1}{u}= -frac{1}{f}+frac{1}{v}Rightarrow frac{1}{f}=frac{1}{v}-frac{1}{u} 

Hence the Lens formula

Similarly we can prove the lens formula for a concave lens.

Illustration: An object is placed 30 cm from a convex lens. A real image is formed 20 cm from the lens. Find the focal length of the lens.

Solution: By convention, the object is placed to the left of the lens. Hence, u is negative. Since the image is real, the transmitted rays actually intersect. The image is thus formed on the right of the lens as shown in figure.

Hence, v is positive. Thus,

        u = - 30 cm       and        v = + 20 cm.

From the lens equation  frac {1}{v}-frac {1}{u}=frac {1}{f}   we have,

                      frac {1}{f}=frac {1}{+20; cm}-frac {1}{-30; cm}=frac {5}{60;cm}

or                   f = 12 cm.

Hence the focal length is 12 cm.

Sample Questions
(More Questions for each concept available in Login)
Question : 1

 The focal length is positive for  _______________ .

Right Option : B
View Explanation
Explanation
Question : 2

Which of the following are correct ?

(a) The lens formula u stands for object distance .

(b) The lens formula v stands for image distance .

(c) The lens formula f stands for focal length.

Right Option : D
View Explanation
Explanation
Question : 3

In the lens f' stands for ._____________.

frac{1}{v}-frac{1}{u}=frac{1}{f}

Right Option : A
View Explanation
Explanation
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