Physics, asked by DiyaBansal, 5 months ago

A concave mirror is used for images formation for different positions of an object. what inferences can be drawn about the following when an object is placed at a distance of 10cm from the pole of a concave mirror of focal length 15cm.
1) position of the image
2)size of the image
3) nature of the image

Answers

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
74

Given

  • Focal length = -15 cm
  • Mirror used = Concave
  • Object Distance = -10 cm

To Find

  • Image Distance
  • Size and the nature of the image

Solution

Mirror Formula :- 1/v + 1/u = 1/f

Size of the image :- M = hᵢ/hₒ= -v/u

Image Distance :

→ 1/v + 1/u = 1/f

→ 1/v + 1/-10 = 1/-15

→ 1/v - 1/10 = - 1/15

→ 1/v = 1/10 - 1/15

→ 1/v = 3/30 - 2/30

→ 1/v = (3-2)/30

→ 1/v = 1/30

→ v = 30 cm

Image Size :

→ -v/u = hᵢ/hₒ

→ -30/-10 = hᵢ/hₒ

→ hᵢ = 3 cm

∴ Thr image is virtual, erect, placed at 30 cm, magnified 3 times with the image size 3 cm

Answered by VinCus
62

Given:-

•A concave mirror is used for images formation for different positions of an object..

•A object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm..

To Prove:-

•Position of the image

•Size of the image

•Nature of the image

Solution:-

•Focal length = - 15 cm

•Object Distance = - 10 cm

To Find Image distance,

•Using formula,

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{ \frac{1}{f}  =   \frac{1}{v}   +  \frac{1}{u}  }}}}}}

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{1}{v}  =   \frac{1}{10}   -   \frac{1}{15}  }}}}}

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{1}{v}  =   \frac{3}{30}   -   \frac{2}{30}  }}}}}

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{1}{v}  =   \frac{1}{30} }}}}}

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{ v = 30 \: cm}}}}}}

To Find Size of image,

•h1 = height of the image

•h2 = height of the object

•Using Formula,

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{ \frac{ - v}{u}  =   \frac{h_</p><p>1}{h_2}}}}}}}

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \frac{ - 30}{10}  =   \frac{h_</p><p>1}{h_2}}}}}}

 \leadsto{ \boxed{ \boxed{ \boxed{ \bold{ \red{h_1 = 3 \: cm }}}}}}

•The image is virtual and erect

•The image distance is 30 cm

•The size of the image is 3 cm

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