Question

An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Solution

For convex lens,

Object distance from the lens, u = –40 cm
Focal length, f = 30 cm
Object size, O = 1.5 cm

Using lens formula

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

We get,

\frac{1}{v} - \frac{1}{- 40} = \frac{1}{30}

\frac{1}{v} = \frac{1}{30} - \frac{1}{40} = \frac{1}{120}

i.e.,

v = 120 cm (for real object)

From relation,
Magnification,

m = - \frac{v}{u}, we get m = - \frac{120}{- 40} = + 3

The image formed by the convex lens becomes object for concave lens at a distance of (120 – 8) = 112 cm on the other side.

For concave lens,

Focal length, f = – 20 cm
Object distance, u = + 112 cm (on the other side)
Image distance, v = ?

Using lens formula, we get

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

Now,    $\frac{1}{v} - \frac{1}{112} = \frac{1}{- 20}$

$\Rightarrow$ $\frac{1}{v} = - \frac{1}{20} + \frac{1}{112} = - \frac{23}{560}$

$v = - \frac{560}{23} cm (for virtual object)$

Using the formula of magnification $m = \frac{v}{u}, we get$
$m = - \frac{560 / 23}{- 112} = - \frac{560}{23} \times \frac{1}{112} = - \frac{5}{23}$

Net magnification,

$m = 3 \times (\frac{- 5}{23}) = - \frac{15}{23} = 0.652 (negative due to virtual image)$

and as,
   $m = \frac{I}{0}$
   $I = m \times 0 = 0.652 \times 1.5 = 0.98 cm (size of final image) .$

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Wave Optics

An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

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