Question

an object, 7 cm in size, is placed at 27 cm in front of a concave mirror of focal length 18 cm. at what distance from the mirror should a screen be placed in order to obtain a sharp image? find the nature and the size of the image.
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Answer 1

Solution:

Given:

⇒ Object size (h) = +7 cm

⇒ Object distance (u) = -27 cm

⇒ Focal length (f) = -18 cm

To Find:

⇒ Image distance (v)

Formula used:

$\sf{\implies \dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}}$

Now, put the value in the formula. We get,

$\sf{\implies \dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}}$

$\sf{\implies \dfrac{1}{-18} = \dfrac{1}{-27} + \dfrac{1}{v}}$

$\sf{\implies \dfrac{1}{-18} +\dfrac{1}{27} =\dfrac{1}{v}}$

$\sf{\implies \dfrac{-3+2}{54}=\dfrac{1}{v}}$

$\sf{\implies \dfrac{1}{v}=-\dfrac{1}{54}}$

${\boxed{\boxed{\red{\sf{\implies v=-54\;cm}}}}}$

The screen should be placed at a distance of 54 cm in front of the concave mirror.

$\sf{\implies m=-\dfrac{v}{u}=\dfrac{h'}{h}}$

$\sf{\implies -\bigg(\dfrac{-54}{-27}\bigg)=\dfrac{h'}{7}}$

$\sf{\implies -2=\dfrac{h'}{7}}$

${\boxed{\boxed{\red{\sf{\implies h'=-14\;cm}}}}}$

Hence, image size = -14 cm.

The image is real, inverted and enlarged in size.

Answer 2

Answer:

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Explanation: