Question

An object 4.0 cm in size is placed at 25.0 cm in front of 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 size of the image

1) v=-37.5 cm, Image is inverted and enlarge

2)v=37.5 cm, Image is erect and diminished

3)v=-12.5 cm, Image is inverted and enlarge

4)v=12.5 cm, Image is erect and diminished

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Answer 1

GIVEN :

• Size of an object is 4.0 cm.
• Object is placed at 25.0 cm in front of a concave mirror.
• Focal length = 15.0 cm.

TO FIND :

• At what distance from the mirror should a screen be placed in order to obtain a sharp image ?
• Nature and size of the image.

FORMULAS USED :

• Distance of the image = $\sf \dfrac {1}{v} \: + \: \dfrac {1}{u} \: = \: \dfrac {1}{f}$
• Size of the image = $\sf \dfrac {-v}{u}$

SOLUTION :

To find the distance of the image,

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

$\implies \sf \dfrac {1}{v} \: + \: \dfrac {1}{-25} \: = \: \dfrac {1}{-15}$

$\implies \sf \dfrac {1}{v} \: = \: \dfrac {1}{-15} \: + \: \dfrac {1}{25}$

$\implies \sf v \: = \: \dfrac {-10 + 6}{150}$

$\implies \sf v \: = \: \dfrac {-4}{150}$

$\implies \sf v \: = \: -37.5 \: cm.$

$\red {\boxed{\sf v \: = \: -37.5 \: cm}}$

$\therefore$ The screen should be placed 37.5 cm in front of lens.

Now,

To find the size of the image,

Size of the image = $\sf \dfrac {-v}{u}$

$\implies \sf h \: = \: \dfrac {-v \times h_0}{u}$

$\implies \sf h \: = \: \dfrac {-(-37.5)\times 4}{-25}$

$\implies \sf h \: = \: \dfrac {-150}{25}$

$\implies \sf h \: = \: - 6 \: cm$

$\red {\boxed{\sf h \: = \: - 6 \: cm}}$

$\therefore$ The image will be inverted.

Because the value of h is negative.

Answer 2

1) v=-37.5 cm, Image is Inverted and enlarge...