A slide with an image 4 cm $\times$ 2 cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. What is the size of the image on the screen?

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Solution

Given u = -10cm
v = 1.1 m
So magnification $m = \frac{v}{u} = \frac{110}{- 10 c m}$
$m = - 11$
hence, size of image will be $44 c m \times 22 c m$ .

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A slide with an image 4 cm × 2 cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. What is the size of the image on the screen?

Given u = -10cm v = 1.1 mSo magnification m=vu=110−10cm m=−11hence, size of image will be 44cm×22cm.

A slide with an image 4 cm $\times$ 2 cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. What is the size of the image on the screen?

Given u = -10cm
v = 1.1 m
So magnification $m = \frac{v}{u} = \frac{110}{- 10 c m}$
$m = - 11$
hence, size of image will be $44 c m \times 22 c m$ .