Question

An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 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

Answer 1

Explanation:

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Answer 2

$\huge\underline\mathrm{Question}$

An object, 4.0. Cm in size, is placed at 25.0cm in front of a concave mirror of focal length 15.0 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???

$\huge\underline\mathrm{AnSweR}$

object-size,h=+4.0cm

Object-distance,u=-25.0 cm;

focal Length,f=-15.0cm;

Image-distance, v=?

Image-size, h=?

we have,

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

$\sf or \: \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{ - 15.0} - \dfrac{1}{ - 25.0} = - \dfrac{1}{15.0} + \dfrac{1}{25.0}$

$\sf or \: \dfrac{1}{v} = \dfrac{ - 5.0 + 3.0}{75.0} = \dfrac{ - 2.0}{75.0}$

$\sf or \: v = - 37.5 \: cm$

$\sf also \: magnification \: m = \dfrac{h}{h} = - \dfrac{v}{u}$

$\sf or \: h = - \dfrac{vh}{u} = \dfrac{(37.5 \: cm) \: ( + 4.0m)}{( - 25.0 \: cm)}$

height of the image,h= -6.0 cm

The image is inverted and enlarged.

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