Math, asked by HarryWalker2036, 15 hours ago

An object 5 cm in size is placed at 12 CM in front of a concave mirror of radius of curvature of 20 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 L .

Answers

Answered by mddilshad11ab
75

Given :-

  • Height of the image {h_(1)} = 5cm
  • The distance of object (u) = - 12cm
  • The radius of curvature (r/2) = 20cm
  • The focal length of the object = -20/2 = - 10cm

To Find :-

  • The nature and size of image = ?

Solution :-

To calculate the nature and size of image at first we have to calculate the distance from the mirror should a screen be placed in order to obtain a image (v). By Applying formula we can calculate (v) then calculate the nature and the size of image.

By using mirror formula :-

1/v + 1/u = 1/f

⇢ 1/v + 1/-12 = 1/-10

⇢ 1/v = -1/10 + 1/12

⇢ 1/v = (-6 + 5)/60

⇢ 1/v = -1/60

v = - 60cm

Therefore, the screen should be placed at 20cm in front of the centre of curvature to obtain the image.

Calculation for size of image :-

⇢ magnification (m) = h_2/h_1 = -v/u

⇢ magnification (m) = h_2/5 = -(-60)/-12

⇢ magnification (m) = h_2 = 5 × 60/-12

⇢ magnification (m) = h_2 = 300/-12

⇢ magnification (m) = h_2 = -25cm

Therefore , the size of the image h_2 = - 25cm

As above you can see that image obtained with negative the sign so, the nature of image be inverted.

Answered by Rudranil420
27

Answer:

Question :-

  • An object 5 cm in size is placed at 12 CM in front of a concave mirror of radius of curvature of 20 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 L .

Given :-

  • An object 5 cm in size is placed at 12 CM in front of a concave mirror of radius of curvature of 20 CM.

Find Out :-

  • 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 L .

Solution :-

Image Distance :

We have :

  • Object Distance = - 12 cm
  • Focal Length = \sf \dfrac{- 20}{2} = - 10 cm

\red{ \boxed{\sf{\dfrac{1}{v} + \dfrac{1}{u} =\: \dfrac{1}{f}}}}

\sf \dfrac{1}{v} + \dfrac{1}{- 12} =\: \dfrac{1}{- 10}

\sf \dfrac{1}{v} =\: \dfrac{- 1}{10} + \dfrac{1}{20}

\sf \dfrac{1}{v} =\: \dfrac{- 2 + 1}{20}

\sf \dfrac{1}{v} =\: \dfrac{- 1}{20}

{\small{\bold{\purple{\underline{v =\: - 20\: cm}}}}}

Size of Image :

\red{ \boxed{\sf{\dfrac{h_2}{h_1} =\: \dfrac{- v}{u}}}}

\sf \dfrac{h_2}{5} =\: \dfrac{- (- 20)}{- 12}

\sf \dfrac{h_2}{5} =\: \dfrac{20}{- 12}

\sf h_2 =\: \dfrac{100}{- 12}

{\small{\bold{\purple{\underline{h_2 =\: - 8.33\: cm}}}}}

Henceforth, the mirror should a screen be placed in order to obtain a sharp image is 20 cm.

The size of the image is 8.33 cm.

The nature of the image is inverted.

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Image Distance :

  • The image distance always equals the object distance.
  • The size of the image is the same as the object (the mirror does not magnify the image).

Object Distance :

  • The distance between the point of incidence and the object placed in front of a mirror is called object distance.
  • The image is the same distance behind the mirror as the object is in front of the mirror.

Focal Length :

  • The focal length ( f ) is the distance from a lens or mirror to the focal point ( F ).
  • This is the distance from a lens or mirror at which parallel light rays will meet.
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