A compound microscope consists of two convex lenses having focal length of 1.5 cm and 5 cm

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When two thin lenses of focal lengths f1 and f2 are kept in contact, prove that their combined focal length 'f' is given by

Consider two thin lenses L1 and L2 of focal lengths f1 and f2 are placed in contact with each other. Suppose that a point object O lies on the principal axis of the two lenses.

shows the formation of the final image I of the object O is two steps. In the first step, the lens L1 produces I’ as the real image of the object O. If C1 O = u and C1 I’ = v’, then for lens L1 ,we have

C2I' ≈ C1I' = +u'

The positive sign has been taken for the reason that the distance of image I’ from C2 is measured in the direction of incident ray. If C2 I = v, then the focal length of lens is given by,

If a single lens of focal length f produces the image I of the object O exactly as the lenses L1 and L2 together do, then

From equation (3), we have

Here, f is called focal length of the equivalent lens.

compound microscope consists of two convex lenses of focal length 2 cm and 5 cm. When an object is kept at a distance of 2.1 cm from the objective, a virtual and magnified image is fonned 25 cm from the eye piece. Calculate the magnifying power of the microscope.

For objective lens (L0 ) u0=-2.1cm

f0 = 2 cm

`1/(f_0)=1/(v_0) -1/(u_0)`

⇒ `1/2 = 1/(v_0) +1/2.1`

⇒ `1/(v_0) = 1/2 - 1/(2.1) = (2.1-2)/(4.2) = 1/42` cm

v0 = 42 cm

Magnifying ower = `(v_0)/(u_0) (1+D/(f_e))`;

[∵ Magnified image is formed 25 cm i.e., least distance ]

Magnifying power = `42/(-2.1)(1+25/5)=-45/(2.1) xx6`

Magnifying power = -120

negative sign shows the final image is inverted

Concept: Optical Instruments - The Microscope

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Text Solution

Solution : D=25cm `u_(0)= -1.6cm, f_(0)= 1.5cm, f_(e )=5cm, M=?` For objective: `-(1)/(u_(0)) + (1)/(v_(0)) = (1)/(f_(0))` `rArr (1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0))` `rArr (1)/(v_(0))= (1)/(1.5) + (1)/(-1.6)` `10 [(1)/(15)- (1)/(16)] = (10)/(15 xx 16)` `rArr v_(0)= (16 xx 15)/(10)= 1.6 xx 15cm = 24cm` Now magnifying power `|m| = - (v_(0))/(|u_(0)|) (1 + (D)/(f e))` `= -(1.6 xx 15)/(1.6) [1 + (25)/(5)]` `= -15 xx 6` `= -90`