Question

An object 4cm in size is placed at 25 cm in front of a concave mirror of focal length 15 cm. At which distance from the mirror should a screen be placed in order to obtain a sharp image ? Find the nature and size of the image.

Answer 1

Answer:

i)At which distance from the mirror should a screen be placed in order to obtain a sharp image ?

=37.5 cm

ii)nature of image->real

Size of image->6cm

Explanation:

Mirror formula 1/f=1/u+1/v

1/15=1/25+1/v

1/v=1/15-1/25

1/v=2/75

v=37.5....positive thus it is real

hi/ho=v/u

hi/4=37.5/25

hi/4=1.5

hi=1.5*4

=6cm

Hope it helps

Please mark as brainliest

Answer 2

Given

• Object size (ho) = + 4cm
• Object distance (u) = - 25cm
• Focal length (f) = -15cm

Find out

• Distance of image
• Size of image
• Nature of image

Solution

★ Apply Mirror Formula ★

$\implies\tt \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f} \\ \\ \\ \implies\tt \dfrac{-1}{25}+\dfrac{1}{v}=\dfrac{-1}{15} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{1}{25}-\dfrac{1}{15} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{3-5}{75} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{-2}{75} \\ \\ \\ \implies\tt v=\cancel\dfrac{-75}{2} \\ \\ \\ \implies\tt v=-37.5cm$

Hence, the distance of image is -37.5cm

★ Magnification ★ = It is the ratio of height or size of image to the height or size of object.

$\implies\tt m=\dfrac{-v}{u}=\dfrac{hi}{ho} \\ \\ \\ \implies\tt \dfrac{-v}{u}=\dfrac{hi}{ho} \\ \\ \\ \implies\tt \dfrac{-(-37.5)}{-25}=\dfrac{hi}{4} \\ \\ \\ \implies\tt 37.5\times{4}=-25\times{hi} \\ \\ \\ \implies\tt 150=-25\:hi \\ \\ \\ \implies\sf hi=\cancel\dfrac{-150}{25}=-6cm$

Hence, the height of image is - 6cm

Hence

• Nature of image
• Real and Inverted