A magnifying lens has a focal length of 10 cm. (i) Where should the object be placed if the image is to be 30 cm from the lens? (ii) What will be the magnification?

Solution

Given , a magnifying lens.
Focal length of the lens, f = 10 cm
Image distance, v = -30 cm

In case of magnifying lens, the lens is convergent and the image is erect, virtual and enlarged and between infinity and the object on the same side of the lens as shown in the above diagram.

Let ‘x’ be the object distance.
Now, using lens formula

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

We have,

\frac{1}{- 30} - \frac{1}{(- x)} = \frac{1}{10}

\Rightarrow

x = 7.5 cm

(ii) Magnification,

m = \frac{I}{O} = \frac{v}{u} = \frac{- 30}{- 8.5} = + 4

Since, the magnification is positive, the image is erect and virtual and four times that of the object.

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Wave Optics

A magnifying lens has a focal length of 10 cm. (i) Where should the object be placed if the image is to be 30 cm from the lens? (ii) What will be the magnification?

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