Question

a 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm at a distance of 12 cm from it used lens formula to determine the position size and nature of the image formed

Answer 1

Distance -36 cm
Height 15 cm
Nature is Virtual And Erect and Enlarged image

Answer 2

Negative sign denotes that the image is formed on the same side as object at 36 cm distance from the optical center.

$h_i=15\ cm$ positive value indicates erect image.

The nature of the image is virtual, erect and magnified.

Explanation:

Given:

• height of the object, $h=5\ cm$

• distance of the object from the lens, $o=12\ cm$

• focal length of the convex lens, $f=18\ cm$

Now, using lens formula:

$\frac{1}{i} +\frac{1}{o} =\frac{1}{f}$

where

i = distance of the image from the optical center of the lens

$\Rightarrow \frac{1}{i} +\frac{1}{12} =\frac{1}{18}$

$\Rightarrow \frac{1}{i} =\frac{1}{36}$

$i=-36$

Negative sign denotes that the image is formed on the same side as object at 36 cm distance from the optical center.

The nature of the image is virtual, erect and magnified.

Now to find the height of image:

$\frac{h_i}{h_o} =-\frac{i}{o}$

$\frac{h_i}{5} =-\frac{(-36)}{12}$

$h_i=15\ cm$ positive value indicates erect image.

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TOPIC: convex lens

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