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12th

Physics

Ray Optics and OPTICAL Instruments

Lens Formula and Magnification

The magnifications produced...

PHYSICS

The magnifications produced by a convex lens for two different positions of an object are m

1

and m

2

respectively (m

1

> m

2

). If d is the distance of separation between the two positions of the object then the focal length of the lens is

HARD

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ANSWER

Given: The magnifications produced by a convex lens for two different positions of an object are m

1

and m

2

respectively (m

1

>m

2

). If d is the distance of separation between the two positions of the object

To FIND the focal length of the lens

Solution:

Separation between object and image for first position,

D=v+u

where u,v are the object distance and image distance, respectively.

So, magnification

m

1

=

u

v

⟹v=um

1

....(i)

When the lens is at second position,

d=v-u

So, m

2

=

v

u

⟹u=vm

2

......(II)

So, m

1

m

2

=1

From lens equation and using eqn (i), we get

f

1

=

v

1

+

u

1

⟹

f

1

=

um

1

1

+

u

1

⟹

f

1

=

um

1

1+m

1

⟹u=

m

1

f(m

1

+1)

.........(iii)

From lens equation and using eqn (ii), we get

f

1

=

v

1

+

u

1

⟹

f

1

=

vm

2

1

+

v

1

⟹

f

1

=

vm

2

1+m

2

⟹v=

m

2

f(m

2

+1)

.........(iv)

substituting the values of eqn(iii) and(iv) in the following equation, we get

d=v−u

⟹d=

m

2

f(m

2

+1)

−

m

1

f(m

1

+1)

⟹d=f(

m

1

m

2

m

1

(m

2

+1)−m

2

(m

1

+1)

)

⟹f=

m

1

m

2

+m

1

−m

1

m

2

−m

2

DM

1

m

2

⟹f=

m

1

−m

2

dm

1

m

2

but m

1

m

2

=1

Therefore the focal length becomes,

f=

m

1

−m

2

d

solution